Foundations of gas dynamics 9781316014288, 1316014282

Table of contents :
Content: 1. Properties and kinetic theory of gases
2. Basic equations and thermodynamics of compressible flow
3. Acoustic wave and flow regime
4. One-dimensional isentropic flow, shock, and expansion waves
5. One-dimensional flows in channels
6. One-dimensional flows with friction
7. One-dimensional flows with heat transfer
8. Equations of multidimensional frictionless flow subject to small perturbation
9. Applications of small perturbation theory
10. Method of characteristics for two independent variables
11. Unsteady flow
12. Introduction to inviseid hypersonic flows.

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Preface

Before embarking on writing this book, I asked myself: With some excellent textbooks already covering this subject, why another one? In one aspect, this book is probably not different from the others in the field, in that it results from over thirty years of experience in learning and teaching the subject of compressible flow (or gas dynamics). The present book is written in what I believe to be an efficient way of learning and presenting the subject materials, reflecting my own perspective as a student and, more importantly, with feedback from students I have taught during the last two decades. Examples and exercise problems, which are many for this level of coverage, are chosen similarly. For a typical mechanical and aerospace engineering curriculum, this book might serve well for senior- or first-year graduate students. The materials are more than sufficient for a one-semester course, allowing the instructor and students to choose materials based on the desired level of coverage. Practicing engineers should also find this book a useful review. One readily finds that the physics of the subject defines the theme of the book. Results from mathematical derivations are explained so that readers can develop physical understanding and intuition. Therefore, I have made no attempt to incorporate numerical methods for solving governing equations of compressible flow. The approach is intuitive and students should only need a simple scientific calculator to follow and understand the materials. Tables and graphs are included wherever necessary to present key results and solve example and exercise problems. Detailed numerical treatises of gas dynamics can be found in many excellent books on this specialized topic, such as those listed in the Reference section. The example problems are chosen to facilitate understanding essential concepts and they might recur in several chapters, even with the same given Mach numbers, wedge and shock wave angles, and so on, to illustrate the development of theories as one progresses from one chapter to another with increasing depth. No attempt is made to obtain third decimal accuracy in the example problems, whether using a table or graph, because effort should be better focused on understanding the materials. While striving to be intuitive and emphasizing physical understanding, the rigor of theory derivation is not lost. Detailed derivations are presented when they best ix Downloaded from https:/www.cambridge.org/core. Columbia University Libraries, on 24 Jun 2017 at 04:19:22, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/9781316014288.001

x

Preface

illustrate the essence of how commonly used gas-dynamic equations are formulated, while teachers can assign the derivations of other equations as exercise problems. Exercise problems are chosen in a similar manner to the example problems (with the goal of enhancing the grasp of the materials). These exercise problems, although not voluminous, should be most beneficial to help with the student’s learning. The support of my family – my wife, Jessie, and my sons, Dar-Wei and Ryan – has been the greatest asset in all my endeavors. The education I received from National Cheng Kung University in Taiwan and the University of Michigan (Ann Arbor) has helped me prepare for my career as a teacher and researcher. I would like to dedicate this book to my family and these two institutions.

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1

Properties and Kinetic Theory of Gases

The properties of gases have direct effects on how gases behave as they undergo mechanical or thermal changes. Thermal changes may result from heating due to heat-releasing chemical reactions, such as those taking place in the chemical propellant in rocket chambers, or from aerodynamic heating as on the external surfaces of high-speed flight vehicles or during reentry to the atmosphere. Mechanical changes are associated with processes such as compression (usually accompanied by deceleration in flowing systems) and expansion (usually accompanied by acceleration in flowing systems). These changes usually occur as a consequence of pressure gradients, changes in volume of a given mass, or the existence of body forces arising from gravitational, electrical, or electromagnetic fields. Mechanical and thermal changes may accompany each other, as in the case of isentropic compression/expansion processes or in a constant-volume heating process. They may also take place simultaneously, as in the case of the chemically reacting flow within a rocket chamber, where heating by heat-releasing chemical reactions occur; these chemical reactions can continue through the convergingdiverging nozzle, where the flow undergoes expansion and acceleration, and beyond the nozzle exit. Chemical changes are not within the scope of this book, except where it is noted. The degree with which thermal changes can be measured depends on the properties of the gas. Gas properties are the average quantities of a large number of gas molecules or atoms, which are in a state of constant motion at temperatures above absolute zero. Under normal conditions (i.e., in the macroscopic or “continuum” limit), gas properties can be easily determined experimentally or from thermodynamic considerations. However, in a rarefied gas, with very low density (as one finds under low-pressure conditions such as in the upper atmosphere), such average properties are not readily defined or obtained. It is therefore desirable to look at the behavior and motion of each individual molecule, that is, at the microscopic level. At the microscopic level, one expects the motion of molecules to be influenced by temperature so as to reflect different thermodynamic properties at the macroscopic level. Gas kinetics (or the kinetic theory of gases) is the study of gas molecules at the microscopic level as influenced by temperature. However, kinetic 1 Downloaded from https:/www.cambridge.org/core. Columbia University Libraries, on 24 Jun 2017 at 06:24:27, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/9781316014288.002

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1 Properties and Kinetic Theory of Gases

theory can be applied to both rarefied and dense gases, with their macroscopic properties reflecting the collective behavior of individual molecules.

1.1 Maxwellian Velocity Distribution Assuming the average spacing between molecules of a given gas species is sufficiently large (larger than the diameter of the molecule so that the gas is “dilute” or the volume taken up by the molecules is negligibly small compared with the volume containing them), molecules do not interact (i.e., do not experience forces from one another) except during collisions. Under such circumstances, the distribution of molecular velocity follows the Maxwellian distribution, for gases in equilibrium states (Vincenti and Kruger, 1975): f ðC1 ; C2 ; C3 Þ ¼


m 3=2 h m  i C12 þ C22 þ C32 exp  2πkT 2kT

(1.1)

where C1 , C2 , and C3 , respectively, are the three velocity components in the Cartesian coordinate system and m, k, and T are the mass of the molecule, Boltzmann constant, and the absolute temperature, respectively. The magnitudes of C1 , C2 , and C3 range from –∞ to +∞. Equation (1.1) indicates that f ðC1 ; C2 ; C3 Þ is an even function offfi C1 , C2 , and C3 . When only the magnitude of molecular speed, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u (= C12 þ C22 þ C32 ), is of interest, the probability of finding molecules with speed between u and u þ du (regardless of direction) is  
m 3=2 mu2 u2 exp  (1.2) χðuÞ ¼ 4π 2πkT 2kT where the magnitude ranges from 0 to +∞. The result of Eqn. (1.2) for χðuÞ vs. u for a family of temperatures T is schematically shown in Fig.1.1. It is seen that for realistic temperatures, the value of χðuÞ is less than 1. This also holds true for f ðC1 ; C2 ; C3 Þ, for the reason provided in the following. Using the definite integral and by noting that f is an even function of C1 , C2 , and C3 , ð∞   1
π1=2 exp ax2 dx ¼ 2 a 0 one finds ð þ∞ f ðC1 ; C2 ; C3 ÞdC1 dC2 dC3 ∞       ð∞ ð∞
m 3=2 ð ∞ mC12 mC32 mC22 3 ¼2 exp  dC1 exp  dC2 exp  dC13 2πkT 2kT 2kT 2kT 0 0 "0 # 3
m 3=2 1 2πkT 1=2 ¼ 23 ¼1 2πkT 2 m

(1.3)

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χ (u)

1.2 Characteristic Molecular Velocities

T3 > T2 > T3

Figure 1.1 Schematic of probability function χðuÞ of finding a molecule having a speed in the range of ðu; u þ duÞ, as function of absolute qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi temperature; u ¼ C12 þ C22 þ C32 is

T1 T2

3

T3

the molecular speed. Note that the probability of finding high-speed molecules increases with the temperature. Molecular speed, u = √C12 + C22 + C32

u

From Eqn. (1.3) the Maxwellian velocity distribution function, Eqn. (1.1), is thus the probability of finding molecules that have three velocity components between C1 and C1 þ dC1 , C2 and C2 þ dC2 , and C3 and C3 þ dC3 . This probability is also the fraction (which is less than 1) of the molecules that possess the velocity components in the said ranges. The total probability of finding molecules with velocity magnitudes between –∞ and +∞ is 1, as expected, and the probability of finding molecules in any given velocity range is always smaller than 1. The same result of the unity total probability using Eqn. (1.2) can easily be found and is left as an end-of-chapter problem.

1.2 Characteristic Molecular Velocities The Maxwellian velocity distribution enables the calculation of magnitudes of some characteristic velocities of a large number of molecules. Examples include the following. 1. The most probable speed of the molecules (ump ) is the speed at which χðuÞ has the maximum value. By setting the derivative of χðuÞ to be zero, ump is found to be ump ¼

2kT m

1=2 :

2. The average speed (u) is   ð∞
m 3=2 ð ∞ mu2 3 u ¼ uχðuÞdu ¼ 4π u exp  du: 2πkT 2kT 0 0

(1.4)

(1.5)

By adopting the definite integral ð∞ 1 x3 expðax2 Þdx ¼ 2 ; 2a 0 one finds

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1 Properties and Kinetic Theory of Gases



8kT 1=2 ≈ 1:13ump : (1.6) mπ
1=2 3. The root-mean-square speed (urms ¼ u2 ) is related to the kinetic energy of the molecules. u2 is obtained as follows:   ð∞
m 3=2 ð ∞ mu2 u2 ¼ u2 χðuÞdu ¼ 4π u4 exp  du: (1.7) 2πkT 2kT 0 0 u¼

By using the definite integral

ð∞ 3
π 1=2 3kT 4 2 x expðax Þdx ¼ ¼ ; 8 a5 m 0 the root-mean-square speed is urms ¼



3kT 1=2 ≈ 1:22ump : m

(1.8)

The average kinetic energy (or the translational energy) of a molecule is therefore 1 3 eetr ¼ mu2rms ¼ kT 2 2

(1.9)

Finding the values of ump , u, urms , and eetr is straightforward and Problems 1.2 and 1.3 should serve as good examples for such an exercise. A few observations can be made regarding the Maxwellian velocity distribution and the magnitudes of characteristic molecular velocities shown in Eqn. (1.5) through (1.8). (i) The probabilities for molecules to move with the same velocity magnitude but in opposite directions are equal, as expected by examining Eqn. (1.1) as f ðC1 ; C2 ; C3 Þ is an even function of each of the three velocity components. To find the velocity magnitudes, χðuÞ can therefore be used in place of f ðC1 ; C2 ; C3 Þ. (ii) The result of urms > u > ump can readily be understood by considering, for example, the distribution shown in Fig. 1.1 and the weighting factors of u3 and u4 in Eqns. (1.5) and (1.7), where the increasing exponent favors the contribution from high-speed molecules. (iii) These characteristic speeds (urms , u, and ump ) are of the same order of magnitude. Because the speed of sound (denoted by a) results from molecular motion and collisions, one can expect it to be of the same order of magnitude as these three characteristic molecular speeds. Because pressure results from molecular collisions, it is also expected to be closely related to the magnitude of these characteristic velocities, and specifically to urms , as shown in the following section.

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1.3 Pressure and the Ideal Gas Law

5

z

Wall

V2 ϕr ϕi

L V

V1 L

x

L

L

y

Figure 1.2 Collision of a particle (for example, a gas molecule) with walls (left figure) and ~ in a cubic box with length L on each side (right figure). a particle having a velocity of V

1.3 Pressure and the Ideal Gas Law Pressure (p) is defined as the force normal to the surface divided by the surface area on which it exerts; it is therefore a normal stress. According to Newton’s second law of motion, the force exerted by a gas molecule (denoted as the ith molecule) on a surface is the rate of momentum change due to its collision with the surface. Referring to Fig.1.2, assume that the collision is “elastic” (i.e., the angle of reflection equals the angle of incidence, φr ¼ φi , and the speed does not change as a result of ~ r j ¼ jV ~ i j ¼ ui ), that, and that in a small region near the surface, the collision, ur ¼ jV molecules collide only with the surface and not with any other molecule. Let L denote the average distance the molecule travels during collision with the wall, a box of a volume L3 can be constructed with each of six surface areas being equal to L2. Let the magnitude of the velocity normal to the surface be ui;x . Then the momentum change throughout the collision process is 2mi ux and the time to complete the process is 2L=ui;x . Therefore, the pressure due to this molecular collision using Newton’s second law is ð2mi ui;x Þ  ð2L=ui;x Þ  L2 ¼

mi u2i;x mi u2i;x ¼ : L –V

(1.10)

where – V ¼ L3 is the volume of the box. Therefore the pressure on the surface facing the x-direction is Px ¼

1X m u2 i i i;x V –

(1.11)

In a more general form, the pressure in the volume of L3 is taken to be the average of the pressures in the three coordinate directions: p¼

   1 1 X
1 X 2 2 2 px þ p y þ pz ¼ (1.12) m u þ m u þ m u mi u2i i i i i;x i;y i;z ¼ i i 3 3– V 3–V

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1 Properties and Kinetic Theory of Gases

Equation (1.12) can be applied to a mixtures of different species (i.e., mi ≠ mj , where i and j denote different species). For a single gas species, with mi ¼ mj ¼ m. Comparing Eqns. (1.8) and (1.12) and keeping in mind that ui represent all possible velocity magnitudes ranging from 0 to þ∞ leads to

 1 X 2X 1 2X Nmu2rms 2 2 e m u u ¼ m ¼ p¼ e ¼ (1.13) i i tr;i i i i 3– V 3–V i 2 3–V i 3–V where N is the total number of molecules in the volume. By substituting Eqn. (1.9) into Eqn. (1.13), one finds p–V ¼ NkT

(1.14)

The assumptions of large spacing and non-interaction between molecules except during the collision process are essentially the same as the ideal gas assumption of classical, macroscopic thermodynamics. Therefore Eqn. (1.14) is identical to the equation of state for an ideal gas from classical thermodynamics: ˆ p–V ¼ nRT

(1.15)

where n and Rˆ are, respectively, the number of moles of gases contained in volume –V and the universal gas constant (Rˆ ¼ 8:3143 kJ/kmol·K). Setting Nˆ to be the Avogadro’s number (6.02252 × 1023 molecules per mole or 6.02252 × 1023/mole), n ¼ N=Nˆ and k¼

Rˆ Nˆ

(1.16)

which yields the Boltzmann constant k = 1.3805 × 10−23 J/K. Equations (1.14) and (1.15) are called the ideal gas law, valid for gas in equilibrium. Alternatively, gas that behaves according to these equations are called the ideal gas for the very fact that the size of the molecule and the intermolecular distance do not play a role. Because a molecule has a finite size, the ideal gas law would not accurately describe the gas behavior under the circumstance where the intermolecular distance is comparable to the molecular diameter. The non-ideality typically occurs under high pressures, for which the gas density is high and the intermolecular forces becomes important even between collisions, violating the “dilute” requirement for the law’s derivation. At low temperatures, non-ideality also occurs because the molecular kinetic energy is small relative to the potential energy between molecules, leading to relatively more important molecular interactions due to potential field.

1.4 Forms of Ideal Gas Law If a specific gases is of interest, Eqn. (1.14) can be rewritten as P– V¼

M ˆ RT ˆ M

(1.17a)

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1.5 Temperature and Energy

7

ˆ are, respectively, the mass of the gas in the volume and the molar where M and M mass of the gas. Equation (1.15) then becomes P– V ¼ MRT

(1.17b)

ˆ M ˆ being the specific gas constant. Because different gas species have with R ¼ R= different molar masses, their values of R are accordingly different. A table (Table A.1) listing values of R for commonly encountered gases is provided in Table B.1 in Appendix B. Recognizing that density ρ ¼ M=–V and that the specific volume v ¼ 1=ρ, the ideal gas law can assume the following forms: – p ¼ ρRT

(1.17c)

p– v ¼ RT

(1.17d)

and

For a given n, the mass M in the volume (and the associated quantities such as ρ and v) is gas-specific, leading to the associated use of the specific constant, R, as in Eqn. – (1.17b) – (1.17d). If the number of moles (n) or of molecules (N) is independent of the specific type of gas (or they are “universal”), then universal constants k and Rˆ are used, as in Eqns. (1.14) and (1.15). The two forms of ideal gas law, Eqns. (1.14) and (1.17) are identical for R ¼ k=m, thus providing a feasible kinetic explanation of a classical thermodynamic result.

1.5 Temperature and Energy Equation (1.9) indicates that temperature and the average kinetic energy of molecules are related. Since temperature can only be measured in regions containing a large number of molecules, it is appropriate to consider the total kinetic energy of   P 1 P 1 2 2 the molecules in the volume, i 2 mi ui . With i 2 mi ui ¼ Etr , Eqn. (1.13) becomes 2 P– V ¼ Etr 3

(1.18)

which relates the macroscopic P and – V to Etr , with Etr determined from the microscopic behavior of the molecules described by the Maxwellian velocity distribution. By comparing Eqn. (1.18) with Eqns. (1.15) and (1.17), the temperature, T, is directly proportional to Etr . Several observations can be made from these results: (i) The term PV represents a form of energy, which is directly proportional to Etr , the total kinetic energy in the volume (or system). A large amount of Etr , or PV in the classical thermodynamics, can be used to perform a large amount of work, such as filling up automobile tires or driving machine tools. (ii) For a given volume containing a given number of molecules (and a given mass), the system pressure is directly proportional to Etr . This is so because during the collision process, the more energetic molecules (i.e., with high speeds) not only

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1 Properties and Kinetic Theory of Gases

experience larger change in momentum (2mi ui;x in Eqn. [1.10]) but also strike the surface with larger frequencies (i.e., within shorter times, as given by 2L=ui;x in Eqn. [1.10]). (iii) According to Eqns. (1.9) and (1.18), the kinetic energy of a molecule is also directly proportional to temperature, leading to the conclusion that the system pressure is directly proportional to temperature for a fixed volume. This conclusion was arrived at by empirical deduction in classical thermodynamics without examining the velocity distribution and the collisions or molecules, in the form of Eqn. (1.15) or (1.17). Since there is no preferred direction of molecular movement and eetr ¼ 12 mu2rms ¼ 32 kT ¼ 12 m½C12 þ C22 þ C32 , a case exists where 1 1 1 1 mC12 ¼ mC22 ¼ mC32 ¼ kT: 2 2 2 2

(1.19)

This case satisfies the principle (or theorem) of equipartition of energy, which states that for any different mode of the average molecular energy that can be expressed as the sum of quadratic terms, all such terms are equal and each makes a contribution of 12 kT. The principle of equipartition of energy applies to other modes of energy, such as rotation and vibration of molecules, as well. Depending on the structure of the molecule, each of the vibrational and rotational modes has a different number of such terms. The number of these terms of energy is also called the degree of freedom (dof). Let dof be denoted by the symbol ξ. The degrees of freedom besides translation is called the internal degrees of freedom (ξ int ). Because ξ ¼ ξ int þ ξ tr ¼ 2 þ 3 ¼ 5. The energy of a molecule is ξ2 kT. A monatomic gas, such as helium and argon, has three translational terms (i.e., in three independent coordinate directions) and possesses no vibrational energy; its rotational energy is too small compared to its kinetic energy because its negligible moment of inertia. It is therefore “structureless” and has ξ int ¼ 0, ξ ¼ ξ tr ¼ 3. A diatomic molecule has a kinetic energy equal to 32 kT, as for monatomic molecules, and a vibrational energy due to the relative velocity between the two atoms along their connecting axis, and two terms of rotational energy due to rotation about the axes that are perpendicular to the connecting axis. Rotational and vibrational energies of molecules also contain quadratic terms of the relative velocity among atoms and angular velocity of rotation, respectively and, according to the principle of equipartition of energy, each of the vibrational and rotational mode contributes 12 kT to the total energy of a molecule. At normal temperatures, the vibrational energy (including both stretching and bending modes) is negligible compared to the other two energy modes. So for a diatomic gas molecules, the molecular energy is 52 kT or 52 RT. It is noted that for determining pressure, only kinetic energy needs to be considered, although a molecule simultaneously possesses multiple modes of energy. Knowing the energy content of molecules, two types of specific heats can be defined: the constant-volume (cv–) and constant-pressure specific heats (cp ).

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1.5 Temperature and Energy

9

The corresponding specific heat ratio γ is defined as γ ≡ cp =cv–. There parameters can be expressed in terms of ξ and R with good accuracy under normal temperatures: ξ cv– ¼ R; 2

cp ¼

ξþ2 R; 2

γ≡

cp ξ þ 2 ¼ ξ cv–

(1.20)

It is often of interest to consider kinetic energy on a per unit mass basis. ˆ M, ˆ By letting etr be the kinetic energy per kilogram and noting Nˆ k ¼ Rˆ and R ¼ R= etr ¼

e 3 Nˆ kT 3 e tr : ¼ RT ¼ ˆ ˆ ˆ 2 M M=N 2

(1.21)

For polyatomic molecules, the accounting of the rotational and vibrational terms and the internal degrees of freedom can become involved, although all of them have three kinetic terms. If the number of atoms in a molecule is N (N > 2), then there are a total of 3N possible degrees of freedom. A simple principle can be followed to approximately determine ξ int at ordinary temperatures. If the molecule is linear (i.e., all atoms form a straight line, such as in CO2), then it has ξ int ¼ ξ rot ¼ ð3N–5Þ; for a nonlinear molecule (e.g., H2O), ξ int ¼ ξ rot ¼ ð3N–6Þ. For further discussion on internal energies, see Vincent and Kruger (1971),Clarke and McChesnney (1964), and Sonntag and Van Wylen (1991). EXAMPLE 1.1

Find the values of cv–, cp, and γ for CO2 and H2O vapor molecules at

25°C. Solution – At this ordinary temperature, vibrational energy is negligible. The CO2 molecule (N = 3) is a linear molecule, similar to a diatomic molecule. It therefore has three terms of kinetic energy, i.e., ξ tr ¼ 3. Therefore, ξ CO2 ¼ ξ tr þ ξ int ¼ 3 þ ð3  3–5Þ ¼ 7. Using Eqn. (1.21), ¼ cv;CO – 2

cv;CO ¼ – 2

ξ CO2 ξ 7 8:3143kJ=kmol ⋅ K kJ Rˆ ¼ 0:6614 R ¼ CO2 ¼ ˆ 44kg=kmol kg ⋅ K 2 2 M CO2 2

ξ CO2 þ 2 ξ CO2 þ 2 Rˆ 9 8:3143kJ=kmol ⋅ K kJ ¼ 0:8503 R¼ ¼ ˆ 2 44kg=kmol kg ⋅K 2 2 M CO2 γ¼

ξþ2 9 ¼ ¼ 1:2857 ξ 7

For comparison, the published values* of cv–, cp , and γ for CO2 at 25°C are, respectively, 0.6529 kJ/kg·K, 0.8418 kJ/kg·K, and 1.2890. The H2O vapor molecule (N = 3) is not a linear molecule with ξ int ¼ ð3  3  6Þ ¼ 3 and ξ tr ¼ 3. Therefore, ξ H2 O ¼ 6. Using Eqn. (1.21) ˆ H O = 18 kg/kmol, and noting that M 2

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1 Properties and Kinetic Theory of Gases

cv;H – 2 O ¼ 1:3857 kJ=kg ⋅ K cp;H2 O ¼ 1:8476 kJ=kg ⋅ K γ ¼ 1:333: The published values* of cv–, cp , and γ for CO2 at 25°C are, respectively, 1.4108 kJ/kg·K, 1.8723 kJ/kg·K, and 1.3270. *These values are taken from R.E. Sonntag and G.J. Van Wylen, Introduction to Thermodynamics, Classical and Statistical, 3rd ed., John Wiley and Sons, 1991. □ Also see Table 1.

1.6 Characteristic Molecular Length and Knudsen Number For a dilute gas, the average spacing between molecules is sufficiently large compared to the size of the molecules. A question arises: how dilute is a gas (or air), for example, under standard atmospheric conditions? Since the molecules are in constant motion, the spacing between them can be chosen to be the average distance traveled by molecules between consecutive collisions. This particular distance is called the mean free path and is denoted by λ. The mean free path is expected to decrease with increases in the cross-sectional area of the molecule πd2 and its number density ρ=m. For collision between like molecules, λ is found to be m λ ¼ pffiffiffi : 2πd2 ρ

(1.22)

There are various ways of determining the effective diameters of molecules. For water and oxygen molecules, the effective diameters are approximately 0.36 nm and 0.4 nm (or 3.6 × 10−10 m and 4.0 × 10−10 m), respectively (CRC Handbook of Chemistry and Physics, 74th ed., 1993). Many other molecules have diameters on the same order of magnitude. For estimation, let the typical diameter of gas molecules be d ≈ 3.8 × 10−10 m and consider air as an example with a molar mass of 28.9 grams. At the standard conditions of 1 atm and 0°C, an ideal gas occupies approximately 22.4 liters (22.4 × 10−3 m3) at the sea level condition. The mean free path is approximately λ ≈ 5:8  108 m Another way of answering the same question is to assume that molecules are not moving (i.e., static) and that 6.02252 × 1023 molecules orderly pack the volume of 22.4 × 10−3 m3. The average spacing between molecules (δ) under this assumption, can be calculated as δ ≈ ðm=ρÞ1=3 and

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1.6 Characteristic Molecular Length and Knudsen Number

11

δ ≈ 3:3  109 m Therefore for air at 1 atm and 0°C, λ : δ : d ≈ 150 : 10 : 1: Thus the mean free path is much greater than the static molecular spacing, which is in turn much larger than the molecular diameter. The kinetic derivation of the ideal gas law based on the dilute model leads to results consistent with classical thermodynamics, which is based on the so-called continuum medium. So air under standard conditions must be sufficiently dilute, but is still a continuum. For a continuum assumption (i.e., such as classical thermodynamics) requires the volume of interest to contain a sufficiently large number of molecules for the measurements and definition of temperature, density, and pressure to be meaningful. Such a requirement is apparent as suggested by Eqn. (1.13), as the value  P V for a small number of molecules would depend on the of the term i mi u2i =3– few velocities possessed by the few molecules incidentally in the volume at the instant of observation. These few discrete velocities might be found anywhere along the curves in Fig. 1.1 depending on the temperature. In the extreme case when there is no molecule in the volume, no temperature can be calculated according to Eqn. (1.14) or (1.15), let alone being observed or measured! Therefore, the ideal gas law is for continuum gases. If the gas is so dilute that the number of molecules in the volume may fluctuate wildly, then a constant density, and therefore a continuum, does not exist in that volume. Under such conditions, the motion of individual gas molecules have to be studied; the field of such study is called rarefied gas dynamics and the motion of molecules constitute a rarefied gas flow. Different flow regimes can be classified by the Knudsen number, Kn, defined as Kn ≡ λ=L, where L is the characteristic length of the system under consideration. For Kn 0) is the entropy increase due to irreversibility and δQ is the heat transfer across system boundaries

Linear momentum

~ MV

~ V

Angular momentum Energy

~ M~ r V

~ ~ r V

E

e

Entropy

S

s

P

~

F i i

P i

0 ð cs

~i T

δQ T

 system

 system

þI

contributions from all the inflows and outflows. For conservation of mass, momentum, and energy, the properties and the results of the rate of change of the system properties are summarized in Table 2.1. Entropy generation is also described in Table 2.1.

2.3 Conservation of Mass (Continuity Equation) Based on the results of the previous section, the conservation of mass (or the continuity) equation becomes ð ð ð ð ∂ ~ or ∂ ~ ~ ⋅ dAÞ ~ ⋅ dAÞ 0¼ ρd– Vþ ðρV ρd–V ¼  ðρV (2.4) ∂t C–V ∂t C–V CS CS Under the steady state condition, ð
ð  ∂ ~ ¼0 ~ ⋅ dA ρd– V¼ ρV ∂t C– V CS requiring that the net mass accumulation is zero within the CV all the time and that whatever the mass flowing into CV must flow out at the same rate. For an unsteady flow, ð ð
 ∂ ~ ≠0 ~ ⋅ dA ρd– V ≠ 0 and ρV ∂t C–V CS However, the mass of the system still remains unchanged at all times and thus Eqn. (2.4). For an incompressible flow, density is not a function of location or time and therefore ð ð ∂ ∂ ρd– V¼ρ d–V ∂t C– ∂t C–V V

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20

2 Equations and Thermodynamics of Compressible Flow

ð d– V ¼ constant and, therefore,

Incompressibility implies C– V

ð


 ~ ¼ 0. ~ ⋅ dA V

CS

The preliminary design for a jet engine is based on a steady state operation at an altitude of 12,000 m, where the air density is 0.312 kg/m3. Assume that both the inlet and exit cross-sections are circular in shape; the inlet and exit diameters are 0.8 m and 0.39 m, respectively. The air speed is 240 m/s and the engine exit speed is 765 m/s (due mainly to thermal expansion by burning the fuel). Also assume that the velocity at both the inlet and the exit is uniform and that the exit gas density at the exit is 0.434 kg/m3.and pressure is same as the ambient. How much fuel in (kg/s) is needed per second? EXAMPLE 2.1

Solution – Mass conservation for this steady state engine (refer to schematic shown in Fig. 2.4) is ð ð
 ∂ ~ ¼  ρ Vi Ai  ṁF;i þ ρ Ve Ae ~ ⋅ dA ρd– Vþ ρV 0¼ i e ∂t C–V CS For steady state operation, ∂ ∂t

ð ρd–V ¼ 0 C– V

Therefore, ṁF;i ¼ ρi Vi Ai þ ρe Ve Ae ¼ 0:312 þ 0:434

kg m π  240  ð0:8 mÞ2 s s 4

kg m π  765  ð0:39 mÞ2 s s 4

ṁF;i ¼ 2:02 kg=s (Also ṁe ¼ 39:66 kg=s and ṁi ¼ 37:64 kg=s.)

□

2.4 Conservation of Momentum: Newton’s Second Law for Fluid Flow The mass (gas particles) flowing through CV carries both momentum and energy ~ substituted into Eqn. (2.2), the conservation of linear with them. With  ¼ V momentum is

ð ð
 X  DΦ ∂ ~ ~ d–V þ ~ ~ ρV ~ ⋅ dA ¼ ¼ ρ V F V (2.5) i i system Dt system ∂t C– V CS ~ d– ~ dm is the momentum of an infinitesimal mass Note that in Eqn. (2.5) ρV V =V inside CV. Equation (2.5) states that all the forces as a whole exerted on the system (the fluid) would cause the time rate of change of momentum of the mass within the ∂ ð ⋅ Þ ¼ 0 and CV plus the net rate of momentum flux out of the CS. In a steady flow, ∂t the forces on the system cause its momentum flux to change between the entrance

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2.4 Conservation of Momentum (a)

(b)

dl

dFe = –pdAe

dF = –pdA

CV

21

dV = dl • dA

dl

dFi = –pdAi

dV = 12 (dAi + dAe)dl

Figure 2.5 Mass flow through and forces acting on a differential control volume (d– V ) and surface (dS), with the differential length being dl.

and the exit of the CV. Therefore in a steady state flow, the system of gas can be accelerated or decelerated, while the CV itself is not. For example, the aircraft jet engine produces the net effect of accelerating the gas from its inlet to exit, which generates forces on the fluid within the engine and the reaction force on the engine (the control volume) is the thrust. In the steady state operation during cruise, the thrust is equal to and is used to overcome the aerodynamic drag, and the CV (the engine and, by extension, the aircraft) does not accelerate. The forces acting on the system include all surface and body forces (denoted by ~ F B , respectively) acting on the system F S and ~   X  X X ~ ~ ~ ¼ þ (2.6) F F F i Bi Si i i i system

system

system

In the absence of electrical or electromagnetic fields and charged gas particles, the other body force of interest is due to gravity. The gravitational force can usually be neglected unless the elevation or gas density differs significantly from one point to the other in the control volume. For ~ F S , all forces acting on the fluid due to contact should be included, including the pressure and viscous shear forces. It is known that among forces due to pressure only the gauge pressure (p  pamb , where pamb is the ambient pressure) on the surface areas at inlets and outlets contributes to ~ F S. Viscous shear forces and those by mechanical devices such as compressors in a jet ~ S . Therefore, engine can be grouped together and designated by R ð  X ~ þR ~ S;system ~ ¼  ðp  pamb Þ ⋅ dA (2.7) F Si i system

CS

where the negative sign for the pressure term is due to the fact that pressure force ~ as shown in Fig. 2.5). acts in the opposite direction of the local surface vector (dA, Equation (2.5) then becomes ð ð ð
 ~ þR ~ ~ S;system ¼ ∂ ~ d–V þ ~ ρV ~ ⋅ dA ρV V (2.8)  ð p  pamb Þ ⋅ dA ∂t C–V CS CS ~ ) and force (P ~ When momentum (MV i F i Þsystem are replaced in Eqn. (2.5) by angular ~ Þ ~ ) and torque (P T moment (M~ rV i system , respectively, the conservation of angular moment (or the angular principle) is established.

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22

2 Equations and Thermodynamics of Compressible Flow

All conditions are the same as those in Example 2.1 with the pressures of gas equal to 19.40 kPa and 304.89 kPa, respectively at the inlet and the exit. What is the thrust force generated by the engine?

EXAMPLE 2.2

Solution – The momentum equation, Eqn. (2.8) ð ð ð
 ~ þR ~ ~ S;system ¼ ∂ ~ d–V þ ~ ρV ~ ⋅ dA ρV  ð p  pamb Þ ⋅ dA V ∂t C– CS V CS with the steady state condition, leads to     RS;system  ½ð p  pamb ÞAe þ ½ð p  pamb ÞAi ¼ ρV 2 A e  ρV 2 A i Note that at the inlet, p ¼ pamb . Therefore,     RS;system ¼ ½ð p  pamb ÞAe þ ρV 2 A e  ρV 2 A i or RS;system ¼ ðpe  pamb ÞAe þ ṁe Ve  ṁi Vi N π kg m ¼ ð304:89  19:40Þ  103 2  ð0:39 mÞ2 þ 39:66  765 m 4 s s kg m  37:64  240 s s RS;system ¼ 64; 446:7 N Since this is the force acting on the system (the exhaust gas), the reaction on the engine is in the opposite direction, in which the aircraft is propelled. Therefore, the thrust force is T ¼ 64; 446:7 N ði:e:; in the negative directionÞ Note – the thrust due to the fuel at the inlet to the engine is assumed to be small. If the fuel is injected in the direction normal to the x-axis, it does not contribute to □ the momentum in the x-direction and does not affect the thrust produced.

2.5 Conservation of Energy: First Law of Thermodynamics The principle of conservation of energy when applied to a system leads to the first law of thermodynamics. When it is applied to the CV, it becomes

ð ð
 DE ∂ ~ ~ ⋅ dA ¼ eρd–V þ e ρV (2.9) Dt system ∂t C– V CS Similar to mass, energy is a scalar that cannot be generated or destroyed in a system. ð ð
 ∂ ~ . The steady state would require that ~ ⋅ dA In general, ∂t eρd– V ¼  e ρV C– V CS ð
 ~ ¼ 0, that is, the net energy accumulation is zero within the CV at ~ ⋅ dA e ρV CS

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2.5 Conservation of Energy

23

all time and that whatever the amount of energy flowing into CV must flow out at the ð ð
 ∂ ~ ≠ 0 and the influx ~ ⋅ dA same rate. For an unsteady flow, ∂t eρd–V ¼  e ρV C– V

CS

and outflux of energy are not equal, resulting in accumulation of energy within the CV. Two forms of energy are of interest: heat and work. For the system, the first law of thermodynamics states that the net balance of heat transfer and work performed across the system boundaries is the amount of energy to be stored in the system. When the time rates of these processes are of interest, the first law is expressed as

_ ¼ DE or (2.10a) Q_  W Dt system δQ  δW ¼ dE

(2.10b)

_ are, respectively, the rates of heat transfer and work across the where Q_ and W system boundaries and Q and W are the amount of heat transfer and work, respectively. The sign conventions for heat and work are: Q_ and Q are positive/negative when the heat transfer is to/from the system (i.e., heat addition/removal is positive/ _ and W are positive/negative when it is done by/on the system. negative) and W Like heat, work is a form of energy and can be envisioned to be taken away from the system when it is expended to perform work (hence the negative sign in Eqn. [2.10]). The energy E consists of internal, kinetic, and potential energies (denoted by U, KE, and PE, respectively) and the energy associated with the fluid having to “push” its way across the control surface (some call this form of energy is sometimes called “flow work”). The fluid gains energy when flowing from a low pressure inlet to an outlet at a higher pressure, like undergoing compression and receiving work, and loses energy when flowing from high to low pressures. As shown in Fig. 2.5a, the pressure force working against the motion of a differential fluid particle (with ~ moving out of CV is d~ ~ which is opposite in direction a volume of d~ l ⋅ dA) F ¼ pdA, to the pressure force. The work expanded (i.e., done) by the fluid particle when it ~ ⋅ d~ F ⋅ d~ l ¼ pdA l, moves the distance of d~ l out of the control volume is dWp ¼ d~ which is positive; it is negative for the inflow. When the fluid particle is tracked from the inlet to the outlet (Fig. 2.5b), this is the amount of energy change in the fluid due ~ ⋅ d~ to the work done by pressure on the boundaries. Because dA l ¼ d–V , on a per mass basis (using dm ¼ ρd– V ), wp ¼

dWp p ¼ ¼ p–v ρ dm

(2.11)

where – v ¼ 1=ρ is the specific volume (volume per unit mass). This amount of pressure work is done on the mass fluxes across the control surface. Summing over ~: the entire control surface and taking the rate of change d~ l=dt ¼ V

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24

2 Equations and Thermodynamics of Compressible Flow

_p¼ W

ð

ð
 p
~ ~  ~ ~ ⋅ dA ρV ⋅ dA or p–v ρV CS ρ CS


 ð ~ ¼ ~ ⋅ dA wp ρV CS

(2.12)

The total amount of work done by pressure can be separated from other forms of work: _ shear þ W_ others _ ¼W _ pþW W

(2.13)

_ shear denotes energy dissipation rate due to shear stresses (pressure is the where W normal stress) and W_ others is work done by other mechanical mechanisms such as a shaft. Then Eqn. (2.9) becomes ð ð
 ~ _ shear  W _ others ¼ ∂ ~ ⋅ dA _ pW eρd–V þ e ρV (2.14) Q_  W ∂t C– V CS Substituting Eqn. (2.12) into Eqn. (2.14) yields ð ð
 ∂ ~ _ _ ~ ⋅ dA _ eρd–V þ ðe þ p–vÞ ρV Q  W shear  W others ¼ ∂t C– V CS

(2.15)

where, on a per unit mass basis, 1 e ¼ u þ V 2 þ gz 2

(2.16)

with u; 12 V 2 , and gz being, respectively, the internal, kinetic, and potential energies per unit mass, g being the magnitude of gravitational acceleration ~ g and z, the elevation in the direction of ~ g. The internal energy u is the molecular energy ~ ) and the bulk discussed in Chapter 1, where knowledge of the bulk gas velocity (V 1 2 kinetic energy 2 V is not required. Inserting Eqn. (2.16) into Eqn. (2.12) leads to


ð ð  1 ~ _ others ¼ ∂ ~ ⋅ dA _ shear  W eρd– Vþ u þ p–v þ V 2 þ gz ρV (2.17) Q_  W ∂t C– 2 V CS The thermodynamic property enthalpy h is defined as h ≡ u þ p–v so that Eqn. (2.17) is also customarily written as


ð ð  ∂ 1 2 ~ _ _ ~ ⋅ dA _ eρd– Vþ h þ V þ gz ρV Q  W shear  W others ¼ ∂t C–V 2 CS

(2.18)

(2.19)

Consider the same conditions of Example 2.2 with the inlet and exit gas temperature equal to 216 K and 1,010 K, respectively. Find the energy content of the fuel (in kJ/kg). Assume that enthalpy is only a function of temperature h ¼ cp T (which will be discussed in more detail later in this chapter) and that values of cp for air and the exhaust gas in the temperature range of interest are approximately 1.0 kJ=kg ⋅ K and 1.1 kJ=kg ⋅ K, respectively.

EXAMPLE 2.3

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2.6 Special Cases of Conservation Equations

25

Solution – The energy added to the system by burning the fuel can be found using the steady-state energy equation. Assume that the velocity and temperature are uniform at both the inlet and exit planes. Then






ð   DE 1 2 1 2 ~ ~ ¼ h þ V þ gz ρV ⋅ dA ¼ ṁ cp T þ V þ gz Dt system 2 2 CS e 

 1  ṁ cp T þ V 2 þ gz 2 i It is reasonable to assume ze ≈ zi . Then






  DE kg kJ 1 kg m2  1; 010 K þ 39:66 ¼ 39:66 1:1  756 Dt system s kg ⋅ K 2 s s




  kg kJ 1 kg m2  216 K þ 37:64  37:64 1:0  240 s kg ⋅ K 2 s s ¼ 22; 864 kJ=s The energy content of the fuel (H) can be estimated to be

DE 46; 181:5 kJ=s kJ ¼ 22; 862:0 H¼  ṁF;i ¼ Dt system 2:02 kg=s kg Comments – (1) A significant portion of the fuel energy is used to overcome irreversibilities such as friction, heat loss from the engine, sudden expansion, and rapid processes such as heat-releasing combustion. The energy content of the fuel thus should be larger than the calculated value. Most hydrocarbon fuels, including aviation fuels, have energy contents approximately 45,000 kJ/kg. (2) The effect of irreversibility will be discussed in more detail in Section 2.7. □

2.6 Special Cases of Conservation Equations For a steady flow with uniform properties at any flow location, the mass conservation equation is reduced to ð
 ~ ¼ 0 and ðρVAÞ ¼ ðρVAÞ ~ ⋅ dA ρV (2.20) e i CS

where the subscripts i and e denote values determined at the inlet and exit surface, respectively. Much more simplified forms of the momentum equation, Eqn. (2.8), and the energy equation, Eqn. (2.19) can be obtained by considering the following conditions: (1) steady flow, (2) the control volume is a thin stream tube having a differential cross-sectional area (dA), as shown in Fig. 2.6, (3) all properties are uniform at the inlet and exit plane, and (4) heat transfer across the CS (i.e., adiabatic

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26

2 Equations and Thermodynamics of Compressible Flow

Streamline

Figure 2.6 Schematic of a control volume bounded by streamlines with uniform properties.

Streamtube

surface), shear work, and other mechanical work are either zero or negligibly small. Under these conditions, Eqn. (2.8) becomes

ð


 1 2 ~ ~ ⋅ dA h þ V þ gz ρV 2

0¼

(2.21)

CS

Because the only inlet and outlet of the flow are located at the ends of the tube, the ~ ¼ dm = constant at both ends. Then, ~ ⋅ dA continuity, Eqn. (2.5), requires that ρV



1 1 h þ V 2 þ gz ¼ h þ V 2 þ gz ¼ constant (2.22) 2 2 e i In the limit where the stream tube consists of only one streamline, Eqn. (2.22) becomes the well-known Bernoulli equation along a streamline. EXAMPLE 2.4

Consider a compressible air flow in the diverging section of a convergingdiverging nozzle (the flow schematic is shown in Fig. E2.4) with adiabatic frictionless wall. At location 1, the gas temperature and pressure are 500 K and 300 kPa, respectively, and the velocity is 539 m/s. The gas temperature and pressure measured at a downstream location, 2, are 358 K and 93 kPa, respectively. What is the gas velocity at location 2? What is the ratio of the crosssectional areas of location 2 to location 1? Assume that the properties are uniform at any location in the nozzle and the value of cp for air is 1.004 kJ=kg ⋅ K. Solution – Because there is no mechanical and shear work (due to the absence of shaft and walls being frictionless), and no heat transfer (adiabatic walls) and the change in elevation can be neglected, Eqn. (2.21) provides a relationship between the gas velocity and temperature:

1

2

Figure E2.4 Flow schematic.

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2.6 Special Cases of Conservation Equations

27

1 1 cp T2 þ V22 ¼ cp T1 þ V12 2 2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
J m2  ð500  358ÞK þ 539 V2 ¼ 2  cp ðT1  T2 Þ þ V12 ¼ 2  1004 kg ⋅ K s ¼ 758 m=s The area ratio according to the continuity equation, Eqn. (2.20), is A2 ρ1 V1 ¼ A1 ρ2 V2 The velocity ratio has been known and the density ratio can be determined by again using the ideal gas law ρ1 p1 T2 300 358  ¼ 2:310 ¼ ¼ 93 500 ρ2 p2 T1 Therefore, A2 ρ1 V1 539 ¼ 1:64 ¼ ¼ 2:310  758 A1 ρ2 V2 Comments – It is noted that the increase in velocity is associated with an increase in area, which is not possible in incompressible flows. However, in compressible flows, such a phenomenon is expected in the supersonic flow regime. As will be discussed in Chapter 3, the flow in this example is supersonic. The local acoustic pffiffiffiffiffiffiffiffiffi speed is a ¼ γRT . So sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 8314 J=kmol ⋅ K m  ð500 K Þ ≈ 450 a1 ¼ γRT1 ¼ 1:4  28:8 kg=kmol s pffiffiffiffiffiffiffiffiffiffiffi a2 ¼ γRT2 ¼

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 8314 J=kmol ⋅ K m  ð358 K Þ ≈ 380 1:4  28:8 kg=kmol s

The Mach number is defined as M ≡ V=a. Therefore, M1 ¼

539 758 ¼ 1:20 and M2 ¼ ≈ 2:0 450 380

At both locations, the Mach number is greater than 1 and the flow is supersonic and accelerates from location 1 to location 2. Note that A2 =A1 > 1; it will be seen in Chapter 4 that an increase in the cross-sectional area for a supersonic flow □ leads to acceleration, contrary to a subsonic flow.

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28

2 Equations and Thermodynamics of Compressible Flow

With the same assumptions as for the above energy equation (frictionless and adiabatic) and requiring that the streamline is a straight line in the x-direction, the momentum equation, Eqn. (2.8), is simplified to ð ð
 ~ ~ ⋅ dA ð p  pamb ÞdA ¼ V ρV (2.23a) RS;system  CS

CS

and     RS;system  ½ð p  pamb ÞAe þ ½ð p  pamb ÞAi ¼ ρV 2 A e  ρV 2 A i

(2.23b)

Since at any instant the system consists of the fluid moving through the CV, RS;system is also the external force on the fluid at a given instant in the control volume and the reaction force is the “propulsive” force on the CV and the structure (e.g., the aircraft, rocket, etc.), as already demonstrated in Example 2.2 (where T ¼ RS;system ). For a constant-area (Ai ¼ Ae ) duct flow with uniform properties and without external forces (RS;system ¼ 0) at any location, p þ ρV 2 ¼ constant:

(2.24)

This form of momentum equation is useful when one-dimensional flows through a normal shock wave and within a constant-area duct with wall friction and heat transfer, as will be shown in Chapters 6 and 7. A supersonic air flow at velocity 760 m/s and a pressure and temperature of 93 kPa and 358 K, respectively, passes through a plane shock wave in a perpendicular direction. As will be discussed in Chapter 4, a shock wave is a strong compression wave, whose thickness is very small (usually on the order of the mean free path). The air pressure and temperature downstream of the shock wave are 418.5 kPa and 603.4 K, respectively. What is the velocity downstream of the shock wave?

EXAMPLE 2.5

Solution – Because the shock wave is thin and is normal to the flow, the constant-area assumption is good. The control volume can be drawn as shown Fig. E2.5. Because the control volume is thin and the flow is one-dimensional, the heat transfer and frictional forces through the edge of the CV are negligible (RS;system ¼ 0). The heat transfer and the work terms are absent in Eqn. (2.19) and therefore

1

2

V1

V2

p1

p2

T1

T2

Figure E2.5 Property changes across the control volume.

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2.7 Second Law of Thermodynamics and Entropy

29

p1 þ ρ1 V12 ¼ p2 þ ρ2 V22 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ρ RT2 p1 T1 2 V2 ¼ ðp1  p2 Þ þ 1 V12 ¼ ðp1  p2 Þ þ V ρ2 ρ2 p2 p2 T2 1 where the ideal gas law for downstream gas is used and R ¼ ð8314 J=kmol ⋅ KÞ=ð28:8 kg=kmolÞ ¼ 288:7 kJ=kg ⋅ K. Thus sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 288:7 kgJ⋅ K  603:4 K   93 358
m2 3 3 2þ   760 93  10  418:5  10 N=m V2 ¼ 418:5 603:4 s 418:5  103 N=m2 ≈ 276 m=s Comments – Since the gas is compressed through a shock wave (one of the subjects of Chapter 4), one expects a smaller speed downstream of the shock wave in the steady state flow.pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Following the comments of Example 2.4, here M2 ¼ pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi ¼ 276= 1:4 and M1 ¼ V1 = γRT1 ¼ V2 = pγRT 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  288  603:4 ¼ 0:56 760= 1:4  288  358 ¼ 2:0, that is a supersonic flow decelerates to be subsonic □ across a normal shock (a topic of Chapter 4).

2.7 Second Law of Thermodynamics and Entropy The second law of thermodynamics addresses whether a prescribed process undergone by the system can possibly take place, even if it satisfies the first law of thermodynamics. It is known that not all processes are possible; for example, a baseball lying on the floor would not jump to the top of a desk without work done by an external source. All processes can be classified into two categories: reversible and irreversible processes. A reversible process is an idealized one in which all details can be reversed and the system restored to its initial state without leaving any effect (or change) on the system or its surroundings. Irreversible processes are those in which some finite changes occur that cannot be removed without external effect or work. Commonly encountered irreversibilities include friction (for example, due to fluid viscosity), heat transfer due to finite temperature differences, mixing of different liquids, combustion reactions, and so on. It takes some effort to separate oxygen and nitrogen in the air, which is the mixture of the two pure gases; energy expended to overcome friction dissipates into the surroundings is practically impossible to recovered. The reversible heat transfer only takes place with zero temperature difference, and thus takes an infinitely long time. Other finite-rate processes are similarly irreversible: the heatreleasing combustion process and the pressure jump across a shock wave within a very short distance and time (as illustrated in Example 2.5). Perhaps naturally occurring events best illustrate irreversibility. For example, water flows from high to lower grounds. To restore it to the high ground, external

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30

2 Equations and Thermodynamics of Compressible Flow

work, such as by pumping, is needed that constitutes an effect on and by the surroundings. Therefore, the second law dictates in what direction a process can or cannot proceed. In terms of entropy, the second law states that the net effect of a thermodynamic process can only increase the combined entropy of the system and its surroundings. The thermodynamic property entropy (S, in J/kg) is defined for a system undergoing a reversible process, as

δQ (2.25) dS ≡ T rev Therefore, ðS2  S1 Þsys ¼

ð2

δQ T rev 1

(2.26)

where δQ is the heat exchange between the system and its surroundings, positive when the heat is transferred to the system and negative out of the system, and the subscripts 1 and 2 denote the initial and the final states, respectively. Keeping in mind that there exists irreversibility in any realistic process, a fixed mass of fluid flowing through the control volume with heat transfer would experience a change in entropy, greater than that due to a reversible process with the same initial and final states: ð2

ð δQ δQ þI > ðS2  S1 Þsys ¼ (2.27) T T rev CS 1 where I is the entropy increase due to irreversibility and is always positive in value. The exact value of I is dependent on the path or the process taken (a reversible process is associated with I ¼ 0). For example, to deliver the same amount of heat transfer, the larger temperature difference between the system and its surroundings, the greater the value is of I. Consequently, the assumption of reversibility of any realistic process is only an idealization with its quality depending on how sudden or finite the changes are. A process that involves no change in entropy is an isentropic process. From Eqns. (2.26) and (2.27), an isentropic process must possess two features: it is both adiabatic (δQ ¼ 0) and reversible (I ¼ 0). A reversible process with heat transfer leads to ðS2  S1 Þsys > 0 (or < 0) if δQ > 0 (or < 0). An irreversible process without heat transfer still leads to an increase in the system entropy, as I > 0. For an irreversible process with heat loss, whether an increase or decrease in S occurs Ð 2 δQ Ð2 would then depend on the relative magnitudes of 1 δQ T and I, as 1 T < 0 and as I > 0. It is therefore possible that the system entropy decreases if there is heat loss from the system to the surroundings. This, however, causes the entropy in the surrounding to increase. On the other hand, heat addition to the system leads to an increase and a decrease in entropy of the system and the surroundings, respectively. The principle of the entropy increase states that the sum of these two changes

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2.8 Isentropic Process

31

in entropy is greater than zero if the heat transfer process is not reversible. The mathematical proof of this principle is shown in the following. Let ΔSsys and ΔSsurr denote the entropy changes of the system and its surroundings, respectively, due to the process. Thus

δQ δQ 1 1 ΔSsys þ ΔSsurr ¼  þ I ¼ δQ  þI (2.28) Tsys Tsurr Tsys Tsurr where δI is the entropy increase due to irreversibility and is ≥ 0. Two, and only two, scenarios of heat transfer are possible: 1. For heat transfer from the system to its surroundings (δQ < 0) to take place, Tsys > Tsurr . 2. For heat transfer from the surroundings to the system (δQ > 0) to take place, Tsurr < Tsys . In either scenario, ΔSsys þ ΔSsurr ≥ 0

(2.29)

with the equal sign for the reversible process (Tsys ¼ Tsurr and I ¼ 0). The total entropy of the universe, consisting of the system and its surroundings, can only increase, although local (i.e., system) entropy may increase or decrease. The universe is a special case of an isolated system; an isolated system is one through whose boundaries there is no mass or energy flow. For an isolated system, ΔSIsolated ¼ I ≥ 0 The entropy balance for the control volume approach is

ð ð ð
 ð δQ_ DS ∂ q_ ~ ¼ ~ ⋅ dA þ I_ ¼ dA þ I_ ¼ sρd– Vþ s ρV Dt system ∂t C– V CS CS T CS T

(2.30)

(2.31)

_ where the rate form of Eqn. (2.27) is used and q_ ¼ Q=A is the heat transfer per unit area per unit time.

2.8 Isentropic Process For a simple compressible substance such as air and most gases, the only reversible work performed on or by the system is in the form of expansion or compression; that is, δW ¼ pd– V (Sonntag and Van Wylen, 1993). By combining the definition of entropy, Eqn. (2.25), and the energy conservation, Eqn. (2.10b), without kinetic and potential energies, TdS  pd– V ¼ dE ¼ dU or

(2.32a)

In terms of intensive properties, Eqn. (2.23a) is rewritten as

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32

2 Equations and Thermodynamics of Compressible Flow

Tds  pd– v ¼ de ¼ du

(2.32b)

Tds ¼ pd– v þ du ¼ dðu þ p–vÞ  –vdp ¼ dh  –vdp

(2.33)

Then

For ideal gases enthalpy (h) and internal energy (u) are both functions only of temperature and their changes with temperature are expressed as dh ¼ cp dT and du ¼ cv–dT

(2.34)

where cp and cv– are the constant-pressure and constant-volume specific heats. Alternatively,



∂h ∂u and cv– ≡ (2.35) cp ≡ ∂T p ∂T v– In general, both cp and cv– are also functions of temperature. Equation (2.33) can be written as vdp or Tds ¼ cv–dT þ pd–v Tds ¼ cp dT  –

(2.36)

By using the ideal gas law, Eqn. (2.36) becomes dp dT ds ¼ cp dT v T  R p or ds ¼ cv– T þ Rd–

v –

(2.37)

If the values of cp and cv– are constant over the temperature range under consideration, then s2  s1 ¼ cp ln

T2 p2 T2 –v2  R ln or s2  s1 ¼ cv– ln þ R ln T1 p1 T1 –v1

(2.38)

Equation (2.38) can be understood in the following manner. From the viewpoint of gas kinetics, entropy is related to the degree of randomness of the molecule in a system. Fast-moving molecules and large volume contribute to higher degrees of randomness as they both make molecules more difficult to locate. As shown in Chapter 1, the molecular speed increases with T 1=2 . Thus, keeping all other properties constant, an increase in temperature leads to an increase in entropy and so does an increase in volume for a given system of molecules. For the same reason, an increase in pressure while keeping temperature constant leads to a decrease in volume and entropy. For an ideal gas undergoing an isentropic process, the changes in p, T, and –v occur in such a way that s2  s1 ¼ 0. Therefore, Eqn. (2.38) leads to p2 ¼ p1



T2 T1

cp =R

v–2 and ¼ –v1

cv–=R T2 T1

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2.8 Isentropic Process

33

Noting that γ¼

cp and R ¼ cp  cv– cv–

(2.39)

The following relationships are obtained cp γ cv– 1 and ¼ ¼ R γ1 R γ1

(2.40)

Therefore for isentropic relationships one finds p2 ¼ p1

γ=ðγ1Þ 1=ðγ1Þ γ T2 –2 ρ1 v T2 p2 ρ ; ¼ ¼ ; ¼ 2 T1 v1 ρ2 – T1 p1 ρ1

(2.41)

Very often the isentropic nature of the process can also be expressed by p–vγ ¼ constant

(2.42)

Those processes that deviate from the idealization can be described by the following expression for easy comparison, p– vn ¼ constant

(2.43)

where n ≠ γ and n ≠ 1; these processes are called polytropic processes (when n = 1 the process is called the isothermal process as the ideal gas law p–v ¼ RT states). The value of n is dependent on the details of process (and path) and is not unique even for a given system of gas. On the other hand, γ is not process-dependent; it is the ratio of gas properties cp and cv–. An calorically or thermally perfect gas is one for which cp and cv– (and thus γ) are constant. An ideal gas that is calorically perfect is also called a perfect gas. Except where noted throughout this book, gasses are assumed to be perfect. One notable case of imperfect gas behavior is that across very strong shock waves in hypersonic flow, where the extremely large temperature rise may produce dissociation species causing values of cp and cv– to change across the shock. A cylinder-piston device containing 1 kg of air is undergoing an expansion process. The initial and the final pressures are 500 kPa and 200 kPa, respectively, and the initial temperature is 327 °C. For air cp ¼ 1:004 kJ=kg ⋅ K. EXAMPLE 2.6

(a) What is the amount of work performed through the isentropic process? What are the final temperature and volume? (b) What is the amount of work for the same pressure condition if the expansion is done through a polytropic process with n = 1.33? What is the heat transfer? What is the final volume?

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34

2 Equations and Thermodynamics of Compressible Flow

Solution – (a) Isentropic process By using the extensive form of the isentropic relationships and the ideal gas law, p– V γ ¼ constant ¼ C and P–V ¼ MRT the expansion work is ð2

W12 ¼

pd– V¼

ð2

1

¼

 p2 –V2γ –V21γ  p1 –V1γ –V11γ C C
1γ 1γ V – d– V ¼  V – ¼ 2 1 Vγ 1γ 1γ 1–

p2 – V2  p1 – V1 MRðT2  T1 Þ ¼ 1γ 1γ T2 ¼ T1



P2 P1

ðγ1Þ=γ

¼



200 kPa 0:4=1:4 ¼ 0:77 500 kPa

T2 = 462 K W12 ¼

1 kg  0:288 kJ=kg ⋅ K  ð462 K  600 K Þ ¼ 99:4 kJ 1  1:4

To find the final volume, Eqn. (2.41) can be rewritten as –2 – V v2 ¼ ¼ V1 v – –1

V1 ¼ –

1=γ p2 200 kPa 1=1:4 ¼ ¼ 1:924 500 kPa p1

MRT1 1 kg  0:288 kJ=kg ⋅ K  600 K ¼ ¼ 0:3456m3 500  kJ=m3 p V2 ¼ 0:6649m3 –

(b) Polytropic process Replacing γ with n, W12 ¼

T2 ¼ T1

MRðT2  T1 Þ T2 and ¼ 1n T1



P2 P1

ðn1Þ=n



ðn1Þ=n P2 200 kPa 0:33=1:33 ¼ ¼ 0:80 500 kPa P1

T2 = 480 K W12 ¼

1 kg  0:288 kJ=kg ⋅ K  ð480 K  600 K Þ ¼ 103:8 kJ 1  1:33

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Problems

35

The first law of thermodynamics for the system is Q12  W12 ¼ E2  E1 ¼ Mcv ðT2  T1 Þ ¼ M

Q12 ¼ W12 þ M

cp ðT2  T1 Þ γ

cp 1:004 kJ=kg ⋅ K ð480 K  600 K Þ ðT2  T1 Þ ¼ 103:8 kJ þ 1 kg  1:4 γ

¼ 17:7 kJ – 2 –v2 V ¼ ¼ –V 1 –v1

1=n p2 200 kPa 1=1:33 ¼ ¼ 1:992 500 kPa p1 V2 ¼ 0:6883 m3 –

Comments – (1) The polytropic process performs slightly more work, 4.4 kJ, than the isentropic process for the same expansion ratio. However, it requires four times the amount of work in the form of heat transfer to the system. In contrast the isentropic process does not require heat transfer to perform work. Because of the heat transfer to the system and the higher final temperature, the polytropic process results in a larger expanded final volume for the same final pressure. (2) for the isentropic process, c kJ=kg ⋅ K ð462 K  600 K Þ ¼ 0, Q12 W12 ¼ M γp ðT2  T1 Þ ¼ 99:4 kJ þ 1 kg0:288 1:4 □ which satisfies the adiabatic requirement for an isentropic process.

Problems Problem 2.1 Consider air (assumed an ideal and perfect gas) in a container undergoing thermodynamic changes. (a) If the volume is reduced by a factor of 3 due through an isentropic process, what is the change (increase or decrease) in entropy? (b) What is the pressure change? (c) If the above volume change takes place under isothermal conditions, what is the entropy change and what is the pressure change? (d) If the volume change takes place under the constant-pressure condition, what is the final temperature? How can you achieve this change? Problem 2.2 A perfectly insulated box was initially divided into two chambers by a diaphragm. One of the chambers was one-third of the total volume and was filled with a perfect gas with specific gas constant R, while the other was a vacuum. The divider was then removed and the system was allowed time to reach an equilibrium state. For the system of perfect gas in the box, what is the change in entropy?

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36

2 Equations and Thermodynamics of Compressible Flow

Problem 2.3 A perfectly insulated box was initially divided by a diaphragm into two chambers containing same gas species. In the initial separated state (state A), one chamber, chamber 1, has volume, pressure, temperature, and mass equal to –V1 , p1 , T1 , and m1 respectively, while the other has –V2 , p2 , T2 , and m2 . The diaphragm is then removed so that the mixing of gas takes place in a reversible manner. The state after the mixing is completed is state B. (a) Find the final pressure, pB , and entropy difference between the two states ðSB  SA Þ. (b) For T1 ¼ T2 ¼ T, m1 ¼ m2 ¼ m (so that mB ¼ 2 m), find pB and ðSB  SA Þ=m and show that ðSB  SA Þ > 0. (c) Show that the result in (a) can be reduced to be identical to that in Problem 2.2 for m1 ¼ 0. Problem 2.4 Within a perfectly insulated box, a football-shaped object was initially suspended from its upper surface by a string. The string was then cut and the object dropped. After the possible bouncing motion had subsided, the center of mass of the object is lowered by a height h. (a) Find the entropy change from entropy change in the gas. (b) Does the inelastic collisions between the object (for example, effects of friction and acoustic waves due to collisions) play a role in the entropy change? Problem 2.5 An air container is initially at a pressure of 1MPa and a temperature of 600 K. A nozzle attached to it discharges its content into the surroundings until the mass in the container is one third of its original value. What are the final temperature and pressure if the process is isentropic? Problem 2.6 A perfect gas in a piston-cylinder device (initial pressure and temperature are p1 and T1 , respectively) undergoes volume change isothermally and reversibly. (a) Find the heat transfer necessary (q12 ) for the specific volume to change from –v1 to –v2 and the entropy change, ðs2  s1 Þ, as a function of p2 =p1 . (b) The above process is followed by heat transfer that brings the system pressure back to p1 , i.e., p3 ¼ p1 , while keeping –v3 ¼ –v2 . Find the heat transfer q23 , T3 , and ðs3  s2 Þ in terms of p2 =p1 . (c) What is the entropy change, ðs3  s1 Þ, expressed in cp, after completing the above two processes? Problem 2.7 A piston-cylinder system, as shown in the figure, contains air (R ¼ 0:287 kJ=kg ⋅ K and cp ¼ 1:005 kJ=kg ⋅ K) is initially at a temperature of 300 K. The cylinder has a cross-sectional area of 0.1 m2 and an initial volume of 0.02 m3 and the piston mass M = 200 kg. The ambient pressure is 1 atmosphere. The cylinder is then submerged in a thermal bath at 500 K. Time is allowed for the equilibrium to be reached with the thermal bath. Assume no friction between the piston and the cylinder wall. Calculate the following: the heat transfer to, the work done by, and the entropy change in, the air in cylinder. Comment on the findings.

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Problems

37

Problem 2.8 Consider a quasi-one-dimensional gas flow in a variable-area channel with frictionless walls. At two known locations, designated by subscripts 1 and 2, the cross-sectional areas and pressures are A1 , A2 , p1 , and p2 . The gas is considered to be ideal with constant cp and cv– (and therefore a constant γ), that is, it is a perfect gas. If the flow between locations 1 and 2 is to be isothermal, find (a) V2 =V1 and (b) heat transfer from location 1 to location 2, q12 .

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3

Acoustic Wave and Flow Regime

This chapter describes the mechanism and the speed with which small disturbances propagate in continuum fluids, especially in gases, where compressible and supersonic flows are likely to occur. Small disturbances, such as human voices, are known to propagate with the speed of sound waves. The mechanism of propagation determines how the disturbance is felt, thus constituting the signaling mechanism. The speed of sound will be determined in terms of macroscopic fluids properties, although the signaling mechanism has a microscopic, or molecular, root.

3.1 Speed of Sound in Compressible Media Chapter 1 discusses some characteristic molecular velocities of gases based on the kinetic theory without needing to know the continuum behavior of gases. Consider molecules adjacent to the surface. These molecules have a preferred direction of motion (i.e., away from the surface) because they rebound from the surface. As a consequence, the molecules present next to this immediate layer also experience more collisions with molecules coming from the direction of the surface than they would if the surface is not present. Such a preferred direction of molecular motion continues into the fluid far away from the surface, with diminishingly smaller effect in that direction. The frequency and intensity of such collisions must be related to some characteristic speed of the molecules, which is a function of the molecular mass and temperature. The fact (or the message) of the presence of the surface propagates in a preferred direction, that is, away from the surface, with that characteristic speed. Recall from Chapter 1 the three characteristic speeds: the most probable (ump ), the average (u), and the root-mean-square (urms ), as Eqns. (1.4), (1.5), and (1.8) show: ump ¼



2 kT 1=2 m

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3.1 Speed of Sound in Compressible Media

u¼

urms ¼

8 kT πm

39

1=2



3 kT 1=2 m

The values of these three velocities depend on the molecular mass (m) and temperature (T). As speculated in Chapter 1, the speed of sound is on the same order of magnitude of these three velocities. However, it is desirable to express the speed of sound in terms of macroscopic (or continuum) properties of the fluids, such as density, temperature, pressure, the degree of compressibility of the medium, and other properties specific to the particular gas (note that k is a universal constant and is not specific to any gas). Recall that to arrive at the ideal gas law Eqn. (1.14) using the kinetic approach, it is implicit that a sufficient number of molecules are contained in a volume. The ideal gas law can be written as, by combining Eqn. (1.14), which is based on molecular velocity distribution, and Eqn. (1.17b), based on macroscopic observations: p– V ¼ NkT

(1.14)

p– V ¼ MRT or p ¼ ρRT

(1.17)

As discussed in Section 1.3 these two forms of ideal gas law are identical for R ¼ k=m. Equation (1.17) includes density, the potential compressibility (p vs. ρ, depending on the nature of the compression process), and the specific gas constant (R instead of k). The gas kinetics theory that leads to the ideal gas law assumes no intermolecular interactions except during collisions and a negligible volume taken by molecules. Therefore, forces such as gravitational and electromagnetic forces are not of concern. At high pressures and low temperatures, when molecules could come into each other’s potential fields, the intermolecular force can be appreciable, constituting the real, non-ideal gas effects. One of the occasions, where the non-ideal gas effect plays a role in the speed of sound in these media, will be seen in the next section. To obtain the speed of sound in terms of macroscopic properties, consider how the signal of an infinitesimal disturbance propagates in a long tube filled with a gas with a piston at one end (shown in Fig. 3.1a). The piston is initially at rest and is given a sudden motion of a constant but infinitesimal constant velocity dV toward the gas. The gas molecules in the immediate vicinity experience not only the presence of the piston surface but also the added velocity dV; the gas thus experiences compression because the collision with molecules freshly rebounding from the surface with a larger velocity leads to a larger pressure. This immediate layer of molecules relays this message of piston motion to the next layer and into the bulk of the gas. Because of the infinitesimal piston motion, this signal of compression (the compression wave) travels

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40

3 Acoustic Wave and Flow Regime

(a)

Front of the acoustic wave caused by the infinitesimal disturbance, dV dV

p + dp T + dT ρ + dρ

Figure 3.1 (a) Acoustic (or sound) wave generated by the motion or disturbance of a piston in a pistoncylinder device; (b) flow and fluid properties seen by an observer traveling with (or “riding on”) the sound wave. The inherently unsteady wave propagation problem of (a) is transformed into a steady state flow problem in (b).

p a T ρ

(b) a – dV p + dp T + dT ρ + dρ

a p T

ρ

x

with the acoustic speed, denoted by a. As illustrated in Fig. 3.1a, the bulk gas behind the wave has infinitesimal increases in pressure (dp), temperature (dT), density (dρ), and velocity (dV, from the initial value of zero). The wave propagation is unsteady in nature. A gas particle initially at rest ahead of the wave will at a later time experience these changes in properties; that is, at a given location, gas properties vary with time. The unsteady problem is inherently more complicated as it involves one more independent variable, that is, time. It can be made tractable by imagining that an observer riding on the sound wave will observe the bulk gas ahead of the wave flow toward the wave with a constant velocity equal to the speed of sound a and leave with a constant speed a  dV, as shown in Fig. 3.1b. By fixing the coordinate system on the traveling wave, the observation becomes steady. That is, besides the velocity, the properties are p; ρ; T, and so on, ahead of the wave and are p þ dp; ρ þ dρ; T þ dT, etc. downstream of the wave, again as shown in Fig. 3.1b. Such a coordinate transformation is called Galilean transformation and is frequently used when dealing with unsteady problem with a constant reference velocity. For a one-dimensional sound wave (in that its thickness is much smaller than the diameter of the tube, or if the portion far away from the wall is examined, where frictional force and heat transfer are negligible) then the Reynolds transport theorem for mass and momentum (Eqns. (2.5) and (2.8)) become, respectively: 0 ¼ ðρ þ dρÞða  dV ÞA  ρaA

(3.1)

ðp  dpÞA þ pdA ¼ ða  dV Þðρ þ dρÞða  dV ÞA  ðaρaAÞ

(3.2)

Note that the surface force the control volume shown in Fig. 3.1b experiences is only due to pressure. By neglecting the second-order term, Eqn. (3.1) is reduced to adρ  ρdV ¼ 0

(3.3)

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3.1 Speed of Sound in Compressible Media

41

By substituting ðρ þ dρÞða  dV ÞA ¼ ρaA into Eqn. (3.2), the momentum equation becomes dp ¼ ρadV

(3.4)

Combining Eqns. (3.3) and (3.4) yields a2 ¼

dp dρ

(3.5)

Apparently, the value of a described by Eqn. (3.5) depends on the process of generating the disturbance. For example, an isothermal process (where p ¼ ρRT= constant would lead to a2 ¼ RT) and an isentropic process (where p–vγ = constant) would lead to different results. Due to the vanishingly small perturbation, heat transfer, and frictional force, the process leading to Eqn. (3.5) is isentropic. Therefore, it is appropriate to express a in terms of a partial derivative while keeping s constant:

∂p 2 (3.6) a ¼ ∂ρ s Because for a perfect gas undergoing isentropic processes, p–vγ = constant, p = constant × ργ , and p ¼ ρRT. Therefore,

∂p γp ¼ γRT (3.7) ¼ constant  γ  ργ1 ¼ pργ  γ  ργ1 ¼ ∂ρ s ρ So the speed of sound is a¼

pffiffiffiffiffiffiffiffiffi γRT

(3.8)

Equation (3.4) indicates that a compression wave (dp > 0) causes dV > 0, which is depicted in Fig. 3.1a; that is, the wave carries the gas behind it along with it. On the other hand, an expansion wave (dp < 0) is generated by withdrawing the piston (dV < 0) away from the gas. While the wave propagates into the bulk gas at rest, it pushes the gas away from it and the gas behind the expansion wave moves along with the piston. For either the compression or expansion wave, the gas between the piston and the wave moves with the cylinder, while that ahead of the wave remains at rest. As long as the amplitude of either wave is small, the propagation is isentropic and the speed of sound is the same for compression and expansion waves; this is left as an end-of-chapter problem (Problem 3.1). The speed of sound is also called the acoustic speed or acoustic velocity. The wave generated by a small disturbance is called a sound wave or acoustic wave. From Eqn. (3.8), the following observations can be made for a sound wave. (1) The speed of sound is specific for a gas at a given temperature. Recall that ˆ M ˆ or k=m, which is specific to a gas as shown in Chapter 1 and that R ¼ R= γ ¼ cp =cv–, with both cp and cv– also specific to the gas.

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42

3 Acoustic Wave and Flow Regime

pffiffiffiffi (2) For a given gas, the speed of sound is proportional to T with T being the temperature ahead of the sound wave. pffiffiffiffiffiffiffiffiffiffiffi (3) Equations (1.4), (1.5), and (1.8) can be rewritten as u ¼ 2 RT , mp pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi u ¼ 8 RT=π, and urms ¼ 3 RT . For air, γ ¼ 1:4 and a < ump < u < urms . For many common gases, γ < 2, such an ordering of molecular and acoustic speeds are also true. (4) The motion of dV < 0 in Fig. 3.2 will also propagate away from the piston, sending out a signal of expansion. In the isentropic limit, the signal also travels pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi with a speed a ¼ ð∂p=∂ρÞs ¼ γRT . The derivation for this result can be done following similar procedures leading to Eqn. (3.6) and is left to the reader as an end-of-chapter problem. In a medium that is not a perfect gas, the speed of sound can be derived using its isentropic compressibility.

1 ∂ρ (3.9) ks ≡ ρ ∂p s Therefore, ffiffiffiffiffiffiffiffiffiffiffiffiffiffi s

ffi sffiffiffiffiffiffiffi ∂p 1 a¼ ¼ ∂ρ s ρks

(3.10)

For liquids and solids the value of the bulk modulus (Ev ) are often reported and it is related to ks by Ev– ¼

1 ks

(3.11)

Therefore, for solids and liquids, sffiffiffiffiffi Ev– a¼ ρ

(3.12)

Because it takes a very large pressure change (or compressive stress, Δp) to cause a small density change (Δρ) in liquids and solids, much larger speeds of sound can be expected than in gases. If the gas in Fig. 3.1a is replaced by a perfectly rigid solid, for which no deformation or change in ρ is appreciable and ks ¼ 0, the speed of sound is infinity, according to Eqn. (3.10). Find the speed of sound in air, water, and stainless steel at 20 °C. The values of Ev– for water and stainless steel are 2:24  109 N=m2 and 160  109 N=m2 , respectively. The water density is 1,000 kg/m3 and the steel density is approximately 8,000 kg/m3.

EXAMPLE 3.1

Solution – Assuming air is a perfect gas,

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3.2 Speed of Sound in Non-Ideal Gases

aair ¼

43

pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi γRT ¼ 1:4  ð288 N ⋅ m=kg ⋅ K Þ  ð293 KÞ ¼ 343:7 m=s

awater

sffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ev– 2:24  109 N=m2 ¼ ¼ ¼ 1; 496:7 m=s ρ 1; 000 kg=m3

sffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ev– 160  109 N=m2 ≈ ass ¼ ¼ 4; 472:1 m=s ρ 8; 000 kg=m3

□

3.2 Speed of Sound in Non-Ideal Gases pffiffiffiffiffiffiffiffiffi In Section 3.2 the speed of sound in perfect gases, a ¼ γRT , is based on the ideal gas law (or the equation of state) p ¼ ρRT and constant specific heats. Behavior of gases at high pressures and low temperatures tend to deviate from the ideal gas law, as the intermolecular forces and molecular volume cannot be neglected. Under these conditions, the gases are “real” in that the ideal gas law becomes an inadequate description of the gas behavior. A measure of the deviation from the ideal state is the compressibility factor, Z: Z ≡ p–v RT

(3.13)

According to this definition, the closer Z is to unity, the closer the gas behaves like an ideal gas. So Z is a measure and says nothing about the “law” a real gas should follow. A number of equations of state have been proposed for the real gas. Two of them are here described. The first is Van der Waals’s equation of state that takes into account effects of intermolecular forces and finite, non-negligible volumes taken by the molecules (Sonntag and Van Wylen, 1993):

 a  (3.14) v  b ¼ RT pþ 2 – v – where a and b are positive constants specific to each gas. For p ≫ a=–v2 and –v ≫ b, Eqn. (3.14) reduces to the ideal gas law. Replacing –v with 1=ρ and after some algebraic manipulations, Eqn. (3.14) becomes p¼

ρRT  aρ2 1  bρ

(3.15)

This expression again reduces to p ¼ ρRT for a ! 0 and b ! 0. Equation (3.6) or (3.10) will still be used for calculating speed of sound, except that ks in Eqn. (3.10) is a species-specific physical property and has to be predetermined. A more general expression is derived in the following.

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44

3 Acoustic Wave and Flow Regime

For a given function z ¼ zð x; yÞ,

ð∂z=∂xÞy ∂y ¼ ∂x z ð∂z=∂yÞx

(3.16)

Similar for s ¼ sð p; ρÞ

∂p ∂ρ



" #" # ð∂s=∂T Þp ð∂T=∂ρÞp ¼ ¼  ð∂s=∂T Þρ ð∂s=∂pÞρ ð∂T=∂pÞρ ð∂s=∂ρÞp

s

By using Eqn. (3.16) once again for the term in the second bracket where T ¼ T ð p; ρÞ,



ð∂s=∂T Þp ∂p ∂p ¼ (3.17) ∂ρ s ð∂s=∂pÞρ ∂ρ T By taking partial derivatives of s in Eqn. (2.37) with respective to T and p while keeping p and ρ fixed respectively, Eqn. (3.17) becomes  





cp =T ∂p ∂p ∂p ¼ ¼γ (3.18) ∂ρ s ∂ρ T ðcv–=T Þ ∂ρ T Therefore, ffiffiffiffiffiffiffiffiffiffiffiffiffiffi s

ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ∂p ∂p a¼ ¼ γ ∂ρ s ∂ρ T

(3.19)

Substituting the Van der Waals’s equation into Eqn. (3.18) yields the speed of sound in a non-ideal gas: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi γRT (3.20) a¼  2  2γaρ 1  bρ An improvement in predicting the speed of sound, particularly for low temperature, is achieved by modifying Van der Waal’s equation as follows: p¼

ρRT aρ2  T 1  bρ

which is Bethelot’s equation of state, so that sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi γRT 2γaρ a¼  2  T 1  bρ

(3.21)

(3.22)

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3.3 Signaling Mechanism

45

3.3 Signaling Mechanism Because small disturbances send out signals traveling at the speed of sound (a), one natural question is how the fluid reacts to the approaching object with V > a. In this scenario, the fluid cannot sense the approaching of the object until being “hit” by the object. The object in this case is said to be traveling with a supersonic speed. Likewise, a flow approaching an object (stationary or moving) with a relative speed that is greater than a is called a supersonic flow; it does not sense and therefore cannot adjust to the presence of the object until it hits the object, when very abrupt changes need to be made (in velocity, density, temperature, and pressure through a shock wave, as will be discussed in details in Chapter 4).

(a)

V=0

V=a

(c) t1

t2

t3

t4

t5

x

a Δt 2a Δt

5 aΔt

(b)

(d)

t1

t2

V>a

μ

a >V>0

t3

t4 t5

x

Zone of influence (action)

Zone of silence (no action)

Figure 3.2 The signaling mechanism of a disturbance moving with a speed V: (a) V ¼ 0, (b) 0 < V < a (i.e., subsonic) (c) the sonic speed, V ¼ a (i.e., sonic) and (d) V > a, (i.e., pffiffiffiffiffiffiffiffiffi supersonic), where a ¼ γRT is the speed of sound. Note that only under supersonic conditions does the distinction between zones of silence (no action) and of influence (action) exists.

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46

3 Acoustic Wave and Flow Regime

It is instructive to examine how acoustic waves generated by moving sources propagate in a quiescent fluid. The results for various speeds ranging from stationary source to sources moving with supersonic speeds are shown in Fig. 3.2. In these figures, the circle with the label 1 and radius 4 aΔt was generated when the object was at location 1, the circle with the label 2 and radius 3 aΔt was generated when the object was at location 2, and so on, while the current location is denoted by 5. For V ¼ 0 (a stationary source), the acoustic wave fronts generated at consecutive time intervals of multiples of an arbitrary length Δt form concentric circles, as shown in Fig. 3.2a. If the object moves with a speed V < a (as illustrated in Fig. 3.2b), then the acoustic wave generated at earlier instant still enclose those generated at later moments, like the case of V ¼ 0. This is because the sound wave travels ahead of the object. However, the wave front spacing is compressed in the direction of the source motion. With V = a, the moving object is catching up with the acoustic wave generated by itself (the wave front spacing is zero in the direction of the source motion), as illustrated in Fig. 3.2c. For V > a, the more recently generated wave fronts intersect the earlier ones, as shown in Fig. 3.2d. In this case, the observer moving with the source would not even hear his/her own voice! For V > a (Fig. 3.2d), lines tangent to the spherical wave fronts can be drawn to form a cone, called Mach cone. This cone has a cone angle, called Mach angle, μ: sinμ ¼

at 1 ¼ Vt M

(3.23)

where M is the Mach number defined as M≡

V a

(3.24)

For M > 1 (Fig. 3.2d), the source is said to be moving with a supersonic speed. The region inside the cone is called the zone of action or the zone of influence. When located outside the cone, the zone of silence, an observer would not feel the disturbance from the source until sometime after the cone moves past the observer. The cone expands with time with a speed of sound in the direction normal to its surface, because the moving source sends out the signal of itself at the speed of sound, even when it is moving with a supersonic speed. It can be said that a concoction such as a supersonic ambulance might be of little use! By using the Galilean coordination transformation for the situation shown in Fig. 3.1b, or in the case of a supersonic flow approaching a stationary object or with a relative speed that is supersonic, the fluid ahead of the object would not sense the existence of the object until the moment it hits the object. Therefore, in flow regime where the relative speed is supersonic, the flow cannot adjust to the object well ahead of the object. If M < 1, the flow is said to be subsonic. In the case of subsonic flow regime, neither a cone nor a cone angle exists, as suggested by Eqn. (3.23) and demonstrated in Fig. 3.2a and 3.2b. A typical subsonic laminar flow over an airfoil is illustrated in

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Problems (a)

(b)

47

Shock wave

Figure 3.3 Streamline patterns over an airfoil of (a) a subsonic flow and (b) a supersonic flow. Note that the streamlines adjust in the upstream region of the airfoil under subsonic conditions while, under supersonic conditions, they are straight until being deflected by the shock wave that forms near the airfoil surface.

Fig. 3.3, where the streamline curvature suggests that the fluid adjusts its path well ahead before reaching the airfoil (the “up-wash” for a lifting airfoil). In other words, the adjustment is made gradually, with changes in the streamline curvature upstream of the airfoil. The gradual adjustment of the fluid in the subsonic regime is not possible in the supersonic regime, as the fluid does not sense that it is approaching the airfoil. Some finite adjustment has to be made at the instant when the fluid “hits” the airfoil. Such a finite and abrupt adjustment (with finite changes in density, pressure, and temperature) is not an isentropic process will be further discussed in the next chapter.

Problems pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Problem 3.1 Referring to Fig. 3.1b, show for a perfect gas that a ¼ ð∂p=∂ρÞs ¼ pffiffiffiffiffiffiffiffiffi γRT if dV < 0 (associated with property changes to p  dp; ρ  dρ; T  dT, etc. downstream of the wave; i.e., the expansion wave still propagates to the right). Problem 3.2 Find the speed of sound using Van der Waals’s equation of state for nitrogen at (a) 101 kPa and 233 K, and (b) 60 atmospheres and 1,200 K. The Van der Waals constants are a ¼ 182:0 N ⋅ m4 =kg2 and b ¼ 1:40  103 m3 =kg. Problem 3.3 Find the speed of sound using Van der Waals’s equation of state for oxygen at 101.3 kPa, and 2,000 K. The Van der Waals constants are a ¼ 136:0 N ⋅ m4 =kg2 and b ¼ 0:995  103 m3 =kg. Problem 3.4 A one-dimensional, small-amplitude pressure pulse (having a magnitude dp) propagates into a quiescent perfect gas at a pressure and temperature of po and To , respectively. Show that the gas velocity and temperature changes are dV ¼

ðγRTo Þ1=2 dp γpo

and dT ¼

ðγ  1ÞTo dp γpo

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48

3 Acoustic Wave and Flow Regime

Problem 3.5 A weak incident pressure wave propagates in quiescent air toward a flat wall with the wave front parallel to the wall (i.e., one-dimensional wave propagation). After reflecting from the wall, the wave travels back into the air, leaving the air behind it quiescent again due to the non-moving boundary condition of the wall. Using Eqn. (3.4) to show that compared to the initially quiescent air, the pressure rise of the air behind the reflected wave is twice that of the air in front of it.

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4

One-Dimensional Isentropic Flow, Shock, and Expansion Waves

Thermodynamic relationships for isentropic processes derived in Chapter 2 are useful simplifications for solving a variety of flow problems. In an isentropic flow, the fluid and flow properties at two locations along a stream line are related by these isentropic relations. In this chapter they are used for flow regimes ranging from subsonic to supersonic. Flow transition from subsonic to supersonic conditions are specifically described in ducts with variable cross-sectional area, with applications to rocket nozzle or other internal flows. Once the possibility for supersonic flow is established, the formation of shock waves becomes a concern. As described in Chapter 3, shock waves are compression waves where the compression process is accomplished over a distance of a few mean free paths. They are therefore very abrupt and thus highly non-isentropic; the stronger the shock wave is (i.e., the larger the compression ratio is), the further the compression process deviates from the ideal isentropic process. Although the flow regimes both upstream and downstream may still be approximated as isentropic, the flow properties across the shock can no longer be related by isentropic thermodynamic relationships. Under isentropic conditions, the energy equation (or energy conservation law) can be used for solution over the entire flow field. With shock waves, changes in flow and fluid properties across them can only be related by solving the three conservation laws (mass, momentum, and energy). To focus on the changes in isentropic flow and due to shock waves, one-dimensional flows of perfect gas are assumed throughout this chapter for simplicity.

4.1 One-Dimensional Isentropic Flow of a Perfect Gas A steady, one-dimensional isentropic flow is a uniform flow with identical constant for all streamlines, for which the Bernoulli equation states

1 p 1 (2.22) h þ V 2 þ gz ¼ u þ þ V 2 þ gz ¼ constant 2 ρ 2

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50

4 One-Dimensional Flow, Shock, and Expansion Waves

Often in the case for compressible flows, changes in enthalpy and the kinetic energy are much larger than that in elevation. Thus between two arbitrary locations on a streamline 1 1 h1 þ V12 ¼ h2 þ V22 2 2 or simply 1 h þ V 2 ¼ constant 2

(4.1)

Although Eqn. (4.1) is derived for an isentropic flow, only the adiabatic condition needs to be satisfied. This is so because both V 2 =2 and h are energies possessed by a fluid particle and are thus not path-dependent. Therefore, it is valid across onedimensional shock waves (further discussed in Section 4.3). Equation (4.1) requires that a change in velocity would lead to changes in h and temperature. In high-speed compressible flow, such changes can be significant. It is desirable to relate these changes to the change in Mach number. Recall that h ¼ cp T and cp γ ¼ R γ1 Therefore, by assuming a constant cp ,



1 V2 γ  1 V2 γ1 2 M ¼ constant ¼T 1þ Tþ ¼T 1þ 2 cp 2 γRT 2

(4.2)

and γ1 T2 1 þ 2 M12 ¼ T1 1 þ γ1 M22 2

(4.3)

By recalling other isentropic relationships, p2 ¼ p1

γ=ðγ1Þ T2 ¼ T1

–2 ρ 1 v ¼ ¼ v1 ρ 2 –

2 1 þ γ1 2 M1

!γ=ðγ1Þ

2 1 þ γ1 2 M2

2 1 þ γ1 2 M1 2 1 þ γ1 2 M2

(4.4)

!1=ðγ1Þ (4.5)

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4.2 Stagnation Properties in an Isentropic Flow

51

4.2 Stagnation Properties in an Isentropic Flow In the above discussion, if at one of the locations (for example location 1) the flow is brought to adiabatically rest (i.e., V = 0), the enthalpy at this location is the stagnation (or total) enthalpy. Because there is no heat transfer, the value of enthalpy remains constant on any given stream line. Letting the subscript “t” denote the stagnation condition, Eqns. (4.1) and (4.2) become 1 ht ¼ cp Tt ¼ h þ V 2 2

(4.6a)

and Tt ¼ T þ





1 V2 γ  1 V2 γ1 2 M ¼T 1þ ¼T 1þ 2 cp 2 γRT 2

(4.6b)

Therefore Tt γ1 2 M ¼ 1þ 2 T

(4.6c)

Equation (4.6c) is valid as long as the flow is adiabatic (even if not isentropic) and the total enthalpy is fixed. For an isentropic flow p–vγ ¼ constant and Eqn. (2.41) becomes pt ¼ p

γ=ðγ1Þ Tt γ  1 2 γ=ðγ1Þ M ¼ 1þ 2 T

(4.7)

ρt ¼ ρ

1=ðγ1Þ Tt γ  1 2 1=ðγ1Þ M ¼ 1þ 2 T

(4.8)

Equations (4.6c) through (4.8) suggest that stagnation properties remain constant on a given stream line and throughout the one-dimensional flow, as the stagnation state can be achieved by bringing the flow from any point to a rest (in an adiabatic manner for Tt and in an isentropic manner for pt and ρt ). Thus they can serve as “reference” states for a stream line. If the flow is uniform and isentropic, then they serve as the reference states throughout the entire flow. Using Eqns. (4.6c) through (4.8), thermodynamic properties (T, p, and ρ) can be determined knowing the value of Mach number M, or vice versa. Numerical results of equations are tabulated in Appendix C as functions of the Mach number for a number of values of γ. For γ ¼ 1:4, these results are plotted in Fig. 4.1, which also contains the Prandtl-Meyer function, , and the area ratio A=A for isentropic channel flows. Both  and A=A are discussed in details in Chapter 5. Consider a uniform steady flow of air in an adiabatic nozzle that has the following properties at location 1: T = 400 K, p = 200 kPa, and V = 400 m/s.

EXAMPLE 4.1

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52

4 One-Dimensional Flow, Shock, and Expansion Waves

102

υ υ

A/A*

101

A/A* T/Tt 100 ρ/ρt T/Tt

p/pt 10–1

ρ/ρt p/pt

10–2 10–2

10–1

100

101

M

Figure 4.1 Graphical representation of isentropic relations among thermodynamic properties (p, ρ, and T) and channel cross-sectional area (A) with Mach number (M). Note that all properties are referred to the reference/stagnation value (with the subscript t), while the reference area is that of the throat of a channel (A ). Note that  is the Prandtl-Meyer function, to be discussed in Chapter 5.

At a downstream location (location 2), the pressure decreases to 150 kPa. What are the velocity and temperature at this location? Solution – Assume that the flow is steady, that there is no heat transfer due to adiabatic nozzle walls, and that air is an ideal gas with constant specific heats over the temperature range of the problem. Because the flow is uniform, there is no shear work and the flow is thus isentropic. Therefore stagnation properties can be used as references throughout the flow. First of all, the Mach number is needed for finding out these reference properties. M1 ¼

pt1 ¼ p1

V1 V1 400 m=s ¼ pffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:996 a1 γRT1 1:4ð288 J=kg ⋅ KÞ  400 K





1:4=ð1:41Þ γ  1 2 γ=ðγ1Þ 1:4  1 2 M1 ½0:996 1þ ¼ 1þ ¼ 1:884 2 2

or pt1 ¼ 376:8 kPa

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4.2 Stagnation Properties in an Isentropic Flow

Tt1 γ1 2 1:4  1 M1 ¼ 1 þ ½0:9962 ¼ 1:198 ¼1þ 2 2 T1

or

53

Tt1 ¼ 479:2 K

For the isentropic flow, pt2 ¼ pt1 and Tt2 ¼ Tt1 . Therefore,



γ  1 2 γ=ðγ1Þ 1:4  1 2 1:4=ð1:41Þ pt2 pt1 p1 M2 M2 1þ ¼ 1þ ¼ 2 2 pt1 p1 p2

200 kPa ¼ ð1Þð1:884Þ ¼ 2:508 150 kPa M2 ¼ 1:226 pt2 ¼ p2

Therefore, T2 ¼

V2 ¼ M2

Tt2 1þ

γ1 2 2 M2

¼

479:2 K 1þ

2 1:41 2 ½1:226

¼ 368:4 K

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi γRT2 ¼ 1:226  1:4ð 288 J=kg ⋅ K Þð368 K Þ ¼ 472:3 m=s

Notes – A pressure decrease is accompanied by an increase in velocity, which is consistent with the results expected from the Bernoulli principle. The value of T2 can also be found by directly using Eqn. (4.3); this is left as an exercise. The ratios are often expressed as static to stagnation values, as shown in Appendix C. □ At low Mach numbers, flow may be assumed to be incompressible. Under such conditions, the difference between stagnation and static values of a given property is small. If a 5% difference is the allowable limit for such an assumption, what would be the Mach number for that limit for air? Is the flow around a car moving at 65-mph (miles per hour) incompressible?

EXAMPLE 4.2

Solution – An examination of Eqn. (4.6c) through (4.8) reveals that the pressure ratio possesses the largest exponent. Therefore, setting pt ¼ p



γ1 2 M 1þ 2

γ=ðγ1Þ

¼ 1:05

M ¼ 0:26 An air speed of 65 mph is equal to V ¼ 28:9 m=s. Under the normal atmospheric condition, T = 293 K and M¼

V 28:9 m=s ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:084 < 0:26 a 1:4ð 288 J=kg ⋅ KÞð293 K Þ

The flow around a 65-mph car can be assumed to be incompressible.

□

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54

4 One-Dimensional Flow, Shock, and Expansion Waves (a) Undisturbed, a a quiescent gas

ΔV

@t1

p

ρ T

χ (b) a Undisturbed, quiescent gas

ΔV

@t2

p

ρ T

χ • • •

(c) @t3

ΔV

a Undisturbed, quiescent gas

p

ρ T

Shock or discontinuity (or jump) in properties

χ

Figure 4.2 (a) Compression waves generated by an infinitesimal velocity of the piston in a piston-cylinder device, at an arbitrary time, t ¼ t1 ; (b) at a later time t ¼ t2 > t1 , the waves generated at later times catch up with those generated earlier; (c) at t ¼ ts > t2 , all the compression waves collapse to form a shock wave. Note that the wave fronts depicted are representative, as there are an infinite number of waves across which the changes in gas properties are infinitesimally small. The piling up of compression waves is due to rises in temperature and the speed of sound caused by compression waves. Across the thin shock wave, the changes in gas properties are abrupt with large gradients, mathematically approximated as infinite.

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4.3 Formation of Normal Shock- and Expansion Waves

55

4.3 Formation of Normal Shock Waves and Expansion Waves If the piston in the piston-tube device, shown in Fig. 3.1a, is given another velocity increment dV to the right, a second sound wave is generated and it propagates to the right following the first sound wave. Because the second wave propagates in the gas that was slightly compressed and heated by the first wave, it has a slightly higher pffiffiffiffiffiffiffiffiffi speed than the first wave, as given by a ¼ γRT , with higher T ahead of the second wave. Given time, the second sound wave will catch up with the first one. One can imagine that a third, a fourth, and more sound waves are generated sequentially, each by increasing the velocity by additional dV to the right over the previous one. If the piston is accelerated from rest to a finite velocity increment (ΔV, which can be viewed as the accumulation of a large number of the infinitesimal increment dV) moving to the right, as sketched in Fig. 4.2a, the change in gas properties is expected to be finite in magnitude (now Δp, ΔT, and Δρ instead of their infinitesimal counterparts across an acoustic wave: dp, dT, and dρ). As the later sound waves eventually overtake those generated earlier, the waves collapse into a very thin region across which finite changes can be observed. The progression toward this collapse of acoustic, infinitesimal acoustic waves are shown in Figs. 4.2b and 4.2c. The collapse of a very large number of infinitesimal compression waves (compression because the piston motion is into the gas) causes a finite increase in pressure across a thin wave. Similar sharp gradients in temperature and density also take place. Since the changes are abrupt, the thin region where they occur constitutes a shock wave. The sharp gradient causes irreversibility and an increase in entropy, as discussed in Chapter 2. Considering the signaling mechanism, the shock wave represents the inability of the gas at rest to sense the piston motion or the arrival of the piston itself; thus the gas could not make infinitesimal adjustment. The time and location of shock formation will be discussed with more details in Chapter 11. If the piston is pulled away from the gas with a constant infinitesimal velocity dV (to the left), as shown in Fig. 4.3, a signal of expansion (i.e., expansion wave) is sent into the quiescent gas to the right of the piston. This expansion wave travels with the pffiffiffiffiffiffiffiffiffi acoustic speed a ¼ γRT , with T being the temperature ahead of the wave as in compression wave propagation. If the acceleration of the piston to the left is from rest to a finite value of ΔV that consists of a large number of infinitesimal magnitude of dV, then the series of expansion waves traveling to the right into the gas at rest will travel with decreasing speeds. This is because a later wave propagates into the gas that now has a lower temperature, and thus a low speed of sound, due to the preceding expansion wave. As a consequence, the spacing between wave fronts increases with time and no “expansion shock” is possible, as shown in Fig. 4.3. For an expansion wave consisting of a series of infinitesimal expansion waves, due to the increasing spacing between the waves fronts, the gradients become increasing smaller with time. The expansion wave can be assumed to be close to being an isentropic process. From the results shown in Fig. 4.2 for the compression wave, the gas at rest assumes a positive absolute velocity once it is downstream of the compression wave,

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56

4 One-Dimensional Flow, Shock, and Expansion Waves (a)

ΔV

@t1

a

Undisturbed, quiescent gas

p

ρ T

χ (b)

ΔV

@t2

a

p

ρ T

χ • • •

(c) @tn Δν

a

p ρ T

χ

Figure 4.3 (a) Expansion waves generated by an infinitesimal velocity of the piston in a piston-cylinder device, at an arbitrary time, t ¼ t1 ; (b) at a later time t ¼ t2 > t1 , the waves generated at later times lack behind those generated earlier; (c) at t ¼ t3 > t2 , all the expansion waves are pulling further away from each other. Note that the wave fronts depicted are representative, as there are an infinite number of waves across which the changes in gas properties are infinitesimally small. The increasing separation between waves is due to the decrease in temperature and the speed of sound caused by expansion waves, so that each later wave propagates with a smaller speed than the previous one and the changes in gas properties are spread over increasing larger distance, resulting in ever decreasing gradients with time. Thus, no shock wave forms, due to expansion.

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4.3 Formation of Normal Shock- and Expansion Waves

57

as seen by an observer in the laboratory. This observation can be intuitively understood, as the piston motion push the gas to the right. In other words, the induced gas motion is in the same direction as the compression wave propagation. It can also be stated that because the piston motion is to the right the gas after being passed by the compression wave (i.e., after receiving the signal of compression action) will have no choice but to move along with the wave to the right due to the motion of the impermeable piston surface. In what direction will be the gas motion downstream of an expansion wave? As illustrated in Fig. 4.3, the expansion wave propagates to the right, even though the piston motion is to the left. To satisfy continuity (i.e., the piston cannot leave a vacuum behind), the gas has to move, after sensing the signal for expansion, to the left. One may then conclude that the gas motion will be in the same direction as the compression wave and the moving normal shock wave, and in the opposite direction to the moving expansion wave. The effect of piston motion described in Fig. 4.3 can be replaced by a diaphragm separating a high-pressure gas and a low-pressure gas, as illustrated in Fig. 4.4a. Assume that the pressure difference is sufficient to cause a shock wave. The diaphragm is ruptured at t ¼ 0þ . At some time later (t > 0), the pressure and the temperature distributions throughout the flow are illustrated in Figs. 4.4b and 4.4c, respectively. The entire flow field now is divided into four regions, with regions 1 and 4 remaining undisturbed at this moment and having pressure and temperature equal to their values at t ¼ 0. The interface between the high- and low-pressure gases, originally represented by the diaphragm before the rupture, acts in similar manner to that of the piston in Fig. 4.3. Upon the diaphragm’s rupture, it simultaneously creates a compression wave propagating in the +x direction and an expansion wave in the −x direction. Because of the need to equalize the pressure across the interface, the action due to compression and expansion is such that p4 > p3 ¼ p2 > p1 . The change from p1 to p2 is abrupt, characteristic of a shock wave, while that from p3 to p4 takes place over a finite distance, as the expansion waves spread out over time. The formation of shock waves is an unsteady process and the shock waves after formation continue to move. Analysis of moving shock waves is presented in Chapter 11. This section is focused on one-dimensional stationary shock waves. The analyses of stationary and moving normal shock waves are similar, except that Galileo coordinate transformation is needed for analyzing the latter, which will be briefly discussed later in this chapter. Examples of normal shock wave include that in a supersonic flow approaching a blunt body, where a normal shock wave forms due to the inability of the flow to sense the presence of the body (as that demonstrated in Fig. 3.3b). Figure 4.5 shows the visualization of the shock wave system formed around a blunt body in a supersonic stream. The portion of the shock wave immediately in front of the nose of the blunt body can be approximately as a normal shock wave, with the flow direction normal to the wave. Away from the nose region, the shock wave assumes the shape of a curve (or a bow). In the bow shock region, the incoming flow makes an angle with the shock wave, thus called oblique shock. A bow shock therefore consists

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58

4 One-Dimensional Flow, Shock, and Expansion Waves (a)

Diaphragm (Interface)

t=0 Quiescent gas p4, T4, ρ4

Expansion waves

(b) t1 > 0 4

Quiescent gas p1, T1, ρ1

Interface

a4 = √γRT4

3

2

Shock a1 = √γRT1

1

p4 (T4) p3 = p2

p2 p1 (T1) χ

T2

T4

T3 T1

Figure 4.4 The shock- and expansion wave system, at t ¼ t1 > 0 and at t ¼ t2 > t1 > 0 (Figures [b] and [c]), established by rupturing the diaphragm that initially, at t ¼ 0 (Figure [a]), separates a high-pressure gas (region 4) from a low-pressure gas (region 1) in a long cylinder. As depicted, the gases in both regions are initially quiescent. Note that the compression propagates into the low-pressure gas, while the expansion wave, into the high-pressure region, and that the gradients in gas properties decrease with time due to the passing of expansion waves.

of a series of oblique shock waves with continuously varying flow angles. The analysis of oblique shock waves is presented later in Section 4.5. This blunt body could be the Pitot-static tube on an aircraft. As an approximation, a portion of shock wave is normal to the flow, as indicated in Fig. 4.5.

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4.3 Formation of Normal Shock- and Expansion Waves

59

(c) Expansion waves

Interface

t2 > t 1 a4 = √γRT4

Shock

3

2

4

1

a1 = √γRT1

p4 (T4) p3 (= p2)

p2 p1 (T1) χ T2

T4

T3 T1 χ

Figure 4.4 (cont.)

Figure 4.5 Shock wave system formed by placing a blunt body in a supersonic stream. (NASA http://www.hq.nasa.gov/pao/History/sp-440/ch6.2hmt)

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60

4 One-Dimensional Flow, Shock, and Expansion Waves

The steady state one-dimensional momentum equation, Eqn. (2.24), can be used to help understand the aforementioned normal shock phenomena: p þ ρV 2 ¼ constant

(2.24)

One-dimensional continuity requires that ρV ¼ constant, and the differential form of Eqn. (2.24), called Euler’s equation, is dp ¼ ρVdV

(4.9)

That is, a positive pressure rise (dp > 0 for a compression or normal shock wave) results in a decrease in the velocity magnitude, or causes the flow to decelerate across the wave. For a traveling compression wave like those depicted in Fig. 4.3, this means that the relative gas velocity relative to the wave decreases after the passage of the compression wave, consistent with the results of Fig. 3.1b. In other words, the gas behind the wave is not leaving the wave with the same speed as its approaching speed, because the gas is also not being pushed to the right by the piston, causing a reduction of speed across a compression/shock wave. To a laboratory observer, the wave thus carries along with it the gas it passes. These results of stationary vs. traveling compression/shock wave are illustrated in Fig. 4.6. For a stationary expansion wave, dp < 0 results in acceleration of the gas flow across it. For a traveling expansion wave, dp < 0 causes an increase in the downstream gas velocity relative to the wave after the passage of the expansion wave. In both cases, the expansion wave “repels” the gas. Figure 4.7 illustrates the differences between observations of a stationary and a traveling expansion waves. The traffic resulting from the change in traffic signals provides good analogies for compression and expansion waves. Consider the steady stream of vehicles approaching a red light at an intersection. The first to sense the signal will have to stop, which in turn sends a stop signal to the immediately following vehicle that will also stop after the first and decrease the space between them. This wave of the stop signal propagates away from the intersection into the still moving vehicles, causing them to decelerate and eventually stop, resulting in an increase in vehicle density, like a compression wave. Some moments later when the red light turns into a green

Add V s V1= 0

–V'1 = –V's

–V'2 2’

1’

(a)

Add (–Vs)

V2 = V s–V2'

Vs

2

1 (b)

Figure 4.6 Schematic demonstrating how an inherently unsteady shock wave propagation problem (part b) can be transformed and treated as a steady state problem by Galileo coordinate transformation (part a). Such a coordinate transformation can be achieved by 0 making observations and measurements by riding on the wave. (Note: V 2 < Vs ; 0 V2 ¼ Vs  V 2 > Vs  Vs ¼ 0.)

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4.4 Governing Equations of Stationary Normal Shock Waves

61

Add V w 2’

1’

V2 = V w–V'2

–V'1 = –Vw

–V'2

Add (–Vw)

2

(a)

Vw

V1= 0

1 (b)

Figure 4.7 Schematic demonstrating how an inherently unsteady expansion wave propagation problem (part b) can be transformed and treated as a steady state problem by Galileo coordinate transformation (part a). Such a coordinate transformation can be achieved 0 0 by making observations and measurements by riding on the wave. (Note: V 2 > V 1 ; 0 0 V2 ¼ Vw  V 2 < Vw  V 1 ¼ 0.)

Control surface (CS) Control volume (CV) 1 V1 M1 p1

ρ1 T1

2 V2 M2 p2

ρ2 T2

Figure 4.8 Schematic of a stationary normal shock wave.

light, the first vehicle will accelerate passing the intersection and away from it. The second and then the third vehicles follow, and so forth, resulting in a wave that sends the signal for motorists to accelerate. In the process of acceleration, the density of vehicles decreases, similar to an expansion wave.

4.4 Governing Equations of Stationary Normal Shock Waves It is desirable to be able to relate fluid properties downstream of the normal shock wave to those in the upstream region. Consider a stationary one-dimensional normal shock wave depicted in Fig. 4.8. Flows both down- and upstream are isentropic and once the relationship between their Mach numbers is found, the isentropic relationships, Eqns. (4.6) through (4.8) can be used. The steady state continuity, Eqn. (2.4), can be rewritten as ρ1 V1 ¼ ρ2 V2

(4.10)

where subscripts 1 and 2 denote the regions upstream and downstream of the shock wave, respectively. Similarly, the steady state momentum equation, Eqn. (2.7), becomes identical to Eqn. (2.24): p1 þ ρ1 V12 ¼ p2 þ ρ1 V22

(4.11a)

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4 One-Dimensional Flow, Shock, and Expansion Waves

For an ideal gas, ρ ¼ p=RT ¼ pγ=γRT. Therefore, ρV 2 ¼ pγM2 and the momentum equation, Eqn. (2.8), becomes     p1 1 þ γM12 ¼ p2 1 þ γM22 (4.11b) For constant values of cp , γ, and R (i.e., perfect gases), h ¼ cp T and cp ¼ γR=ðγ  1Þ. The energy equation, Eqn. (4.22), becomes 1 1 h1 þ V12 ¼ h2 þ V22 2 2

(4.12a)





γ1 2 γ1 2 M1 ¼ T 2 1 þ M2 T1 1 þ 2 2

(4.12b)

and can be rewritten as

Combining the ideal gas law and the definitions of Mach number and speed of sound for perfect gases, the continuity equation, Eqn. (4.10), can be rewritten as pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi p1 p2 M1 γRT1 ¼ M2 γRT2 RT1 RT2 Substituting Eqns. (4.11b) and (4.12b) into Eqn. (4.13) yields rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M12 γ1 2 M22 γ1 2 M1 ¼ M2 1þ 1þ 2 2 2 2 1 þ γM1 1 þ γM2 Squaring both sides of Eqn. (4.14a) produces

 γ1 2 2 2 M12 1 þ γ1 M M 1 þ M 2 2 1 2 2  2 ¼  2 2 2 1 þ γM1 1 þ γM2

(4.13)

(4.14a)

(4.14b)

This equation relates the downstream Mach number M2 to the upstream Mach number M1 by collecting terms containing M2

γ1  γ2 Z þ M24 ð1  2γZÞ  Z ¼ 0 M24 2 where
 2 M12 1 þ γ1 M 1 2 Z¼  2 2 1 þ γM1 Solving the quadratic equation for M22 yields M22 ¼

2 M12 þ γ1 2γ 2 γ1 M1

1

(4.15)

The plot of M2 vs. M1 for γ ¼ 1:4 is shown in Fig. 4.9.

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4.4 Governing Equations of Stationary Normal Shock Waves

63

103 p2 /p1

102

T2/ T1 p2 /p1 T2 / T1

101

ρ2 /ρ1 100

ρ2 /ρ1 M2n pt 2/pt 1

M2n

10–1 10–2

pt 2/pt 1

10–3 10–4 1

M1n

10

Figure 4.9 Property changes across a stationary normal shock wave, as functions of upstream Mach number, M1 for γ ¼ 1:4.

For M1 > 1, M2 < 1 and the flow decelerates across the wave and, according to     Eqn. (4.11b), p2 =p1 ¼ 1 þ γM12 = 1 þ γM22 > 1. Therefore the wave is a compression shock. Equation (4.14b) also has a trivial solution M1 ¼ M2 ¼ 1, which is the limiting case where no flow changes occur across the wave. Conversely for M1 < 1, M2 > 1 and the flow accelerates across the wave and,     according to Eqn. (4.11b), p2 =p1 ¼ 1 þ γM12 = 1 þ γM22 < 1. Therefore the wave is an expansion shock, which is not possible because the expansion leads to decreasing gradients of properties, as discussed in Section 4.3. This impossible scenario for the one-dimensional flow will be later explained further in terms of the entropy change. Note that the above derivation does not assume the existence of a shock wave. An expansion wave is also a possible solution, although an expansion shock is not. As will be demonstrated in Section 4.10, a supersonic flow can continue to accelerate (i.e., expand) by “turning” and, in Section 5.1, by an increase in cross-sectional area in a channel. Now that M2 can be related to M1 , ρ2 , V2 , T2 , and p2 can also be related to ρ1 , V1 , T1 , and p1 , using Eqns. (4.10) through (4.12) and Eqn. (4.15). The following relationships across a stationary normal shock wave can easily be derived. Equation (4.11) is rewritten as p2 1 þ γM12 ¼ p1 1 þ γM22

(4.16a)

Substituting Eqn. (4.15) into this equation yields p2 2γM12 γ  1  ¼ p1 γ þ 1 γ þ 1

(4.16b)

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64

4 One-Dimensional Flow, Shock, and Expansion Waves

Similarly, combining Eqns. (4.12b) and (4.15), the temperature ratio is 

γ1 2γ 2 2 1 þ M M  1 1 1 2 γ1 T2 " # ¼ 2 T1 ð γ þ 1Þ M12 2ðγ  1Þ

(4.16c)

By using the continuity and substituting Eqns. (4.15) for M2 and (4.16c) for the temperature ratio, the density ratio is pffiffiffiffiffiffiffiffiffiffiffi ρ2 V1 M1 γRT1 ðγ þ 1ÞM12 pffiffiffiffiffiffiffiffiffiffiffi ¼ ¼ ¼ (4.16d) ρ1 V2 M2 γRT2 ðγ  1ÞM12 þ 2 Equation (4.16a) shows that the pressure jump across the normal shock wave increases with increasing upstream Mach number M1 , accompanied by a correspondingly smaller downstream Mach number M2 , as seen in Fig. 4.9 for γ ¼ 1:4. For convenience, it is desirable to define the shock strength β, as ðp2  p1 Þ=p1 : β≡

 p2  p1 2γ  2 M1  1 ¼ γþ1 p1

(4.16e)

Equation (4.16e) provides a useful relation for choosing M1 for the desired shock strength:

γþ1 M12 ¼ 1 þ β (4.16f) 2γ Because of the one-dimensional nature of the normal shock wave, there is no heat loss. Therefore, Tt2 ¼ Tt1

(4.17a)

However, the irreversibility due to the finite compression process must lead to changes in some thermodynamic properties such as entropy and stagnation pressure. By using Eqn. (4.12) for static temperatures and the isentropic relationships, Eqns. (4.7) and (4.8), and after performing some algebraic manipulations, Eqn. (2.38) becomes 8 " #γ=ðγ1Þ 9



γ1 γ1 < = 2 2 1 þ M 1 þ M cp s2  s1 c p T2 p2 pt2 1 1 2 2 ¼ ln  ln  ln ¼ ln :pt1 1 þ γ1 M2 ; R R T1 p1 R 1 þ γ1 M2 2

2

2

2

(4.17b) Because cp =R ¼ γ=ðγ  1Þ, this expression becomes s2  s1 pt2 ¼ ln R pt1

(4.17c)

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4.4 Governing Equations of Stationary Normal Shock Waves

65

It is noted from Eqns. (4.15), (4.16a), and (4.17c) that for the trivial case of M2 ¼ M1 ¼ 1, s2 ¼ s1 (i.e., the flow is isentropic) and pt2 ¼ pt1 . Equation (4.16f) requires that for one-dimensional shock (finite pressure jump and β > 0, unlike an isentropic acoustic wave) to occur, it is necessary thatM1 > 1. For an oblique shock (Section 4.6) to occur, it is necessary that M1 > 1 (and M1n > 1 for oblique shocks). In fact M1 in Eqns. (4.16) and (4.17) can simply be replaced by M1n for oblique shocks. For a normal shock, being an irreversible process with large gradients, s2  s1 > 0. The ratio of the stagnation pressure is " #γ=ðγ1Þ " #1=ðγ1Þ γþ1 2 pt2 1 ðs2 s1 Þ=R 2 M1 ¼e ¼ 0, Eqn. (4.17d) requires that pt2 < pt1 . Therefore for a stationary normal shock, the irreversibility leads to a loss in the stagnation (or total) pressure. On the other hand, Eqn. (4.16) requires that p2 > p1 (because it is necessary that M1 > 1), due to the compression effect of the normal shock. The loss in the stagnation pressure means that the ability of the gas to expand and perform the pdV work is reduced. This has important implications when designing supersonic inlets, where normal shock waves are to be avoided. The reason for this is so that the stagnation pressure remains high entering the jet engine, and the capability is preserved for the gas to expand through the turbine stages and the nozzle exit (to perform a large amount of work in driving the compressor and the high-speed exit flow for propulsion). The left-hand side results of Eqns. (4.15) – (4.17) are all functions only of the upstream Mach number M1 and the specific heat ratio γ. They can be conveniently tabulated, in Appendix D, where M1n is adopted to indicate the usefulness of the table for oblique shocks; for normal shocks, M1n ¼ M1 . It is also possible to express the ratios of different properties as functions of each other, rather than as functions of M1 . For example, expressing M1 in terms of the density ratio in Eqn. (4.16c) and substituting it into Eqn. (4.16a) yields

 ρ2 γþ1 γ1 ρ1  1 p2 ¼

 (4.18a) γþ1 p1  ρ2 γ1

ρ1

Alternatively, the density ratio can be expressed as the pressure ratio, as
 γþ1 p2 γ1 p1 þ 1 ρ2
 ¼ γþ1 ρ1 þ p2 γ1

(4.18b)

p1

The result of the Rankine-Hugoniot relation will be discussed in further detail in Section 4.18.

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66

4 One-Dimensional Flow, Shock, and Expansion Waves

An air flow with Mach number 2.0 passes through a normal shock wave immediately ahead of a jet engine inlet. The static air temperature and pressure are, respectively, 216 K and 19.4 kPa. What is the air velocity downstream of the shock? What is the pressure loss? What are the downstream pressure, temperature, and density? Also use the Rankine-Hugoniot relation to find downstream density.

EXAMPLE 4.3

Solution – First of all the Mach number and the sonic speed are needed downstream of the shock wave. For M1 ¼ 2:0, Appendix A provides T1 =Tt1 ¼ 0:5556 while the shock table (Appendix D) provides M2 ¼ 0:5774 and T2 =T1 ¼ 1:6875, pt2 =pt1 ¼ 0:7209, ρ2 =ρ1 ¼ 2:667, and p2 =p1 ¼ 4:50. Therefore, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi V2 ¼ M2 γRT2 ¼ ð0:5774Þ  ð1:4Þð 288 J=kg ⋅ KÞð1:6875  216 K Þ ¼ 221:4 m=s:

γ  1 2 γ=ðγ1Þ M1 pt1 ¼ p1 1 þ ¼ ð19:4 kPaÞð7:8247Þ ¼ 151:80 kPa; 2 pt2 ¼ 0:7209pt1 ¼ 109:43 kPa Δpt ¼ pt2  pt1 ¼ 42:47 kPa The pressure loss is approximately 28%. p2 ¼ 87:3 kPa T2 ¼ 364:5 K Tt2 ¼ Tt1 ¼ ðTt1 =T1 Þ  T1 ¼ 216 K=0:5556 ¼ 388:8 K

ρ2 ¼ 2:667 

p1 19:4  103 kPa ¼ 0:832 kg=m3 ¼ 2:667  ð288 J=kg ⋅ KÞ  ð216 KÞ RT1

By using the Rankine-Hogoniot relation with γ ¼ 1:4
 γþ1 p2 γ1 p1 þ 1 ρ2 6  4:5 þ 1 ¼ 2:667 ¼
 ¼ γþ1 p 6 þ 4:5 ρ1 þ 2 γ1

p1

which is same as using the tabulated value for normal shock.

□

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4.5 Strong and Weak Normal Shock Waves

67

4.5 Strong and Weak Normal Shock Waves For strong shock waves, the shock strength β is large and, according to Eqn. (4.17e), M1 must be large. On the other hand, M1 must be very close to unity for weak shock waves. For strong shock waves, Eqn. (4.15) takes the following form: " 2 # 2 M1 þ γ1 2 M2 ¼ lim 2γ 2 (4.19a) M1 !∞ γ1 M1  1 Therefore, sffiffiffiffiffiffiffiffiffiffiffi γ1 M2 ¼ 2γ

(4.19b)

Therefore, for γ ¼ 1:4, M2 ≈ 0:378 is the minimum value possible if M1 is infinity. As a comparison, use M1 ¼ 10, M2 ¼ 0:388, as seen in Appendix D. After substituting Eqn. (4.19b) into Eqn. (4.16), ratios across the normal shock of other fluid properties can be similarly written as T2 2γðγ  1Þ γ  1 β ≈ M12 ≈ γþ1 T1 ðγ þ 1Þ2

(4.19c)

p2 2γM12 ≈β ≈ p1 γ þ 1

(4.19d)

ρ2 V1 ðγ þ 1Þ ¼ ≈ ρ1 V2 ðγ  1Þ

(4.19e)

   1=ðγ1Þ   pt2 γ þ 1 γ=ðγ1Þ γ þ 1 1=ðγ1Þ 1 ¼ eðs2 s1 Þ=R ≈ ¼ 2 γ1 β pt1 2γM1

(4.19f)

Therefore, as M1 ! ∞, ðT2 =T1 Þ and ðp2 =p1 Þ are linearly related rather than related by the nonlinear relations of Eqns. (4.16b) and (4.16c). Furthermore, ρ2 =ρ1 becomes independent of M1 and pt2 =pt1 eM12=ðγ1Þ , which is eM15 for γ ¼ 1:4. It is not possible for M1 to be infinity, for which the downstream pressure and temperature will also be infinity while the pressure loss is 100%. Above what value of M1 are Eqns. (4.19a) through (4.19f) good approximations?

EXAMPLE 4.4

Solution – pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi One estimation can be obtained by assuming M2 to be 110% of ðγ  1Þ=2γ. In this case for γ ¼ 1:4, M2 ≈ 0:416 and the normal shock table (Appendix D) gives M1 ≈ 4:94. The results with and without the strong shock assumption for fluid properties are

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4 One-Dimensional Flow, Shock, and Expansion Waves

T2 T1 ¼ 5:684 p2 p1 ¼ 28:304 ρ2 ρ1 ¼ 4:980 pt2 pt1 ¼ 0:065

If M2 is 105% of T2 T1 ¼ 10:469 p2 p1 ¼ 57:0 ρ2 ρ1 ¼ 5:444 pt2 pt1 ¼ 0:015

4.745 (strong shock approximation) 28.47 (strong shock approximation) 6.0 (strong shock approximation) 0.122 (strong shock assumption) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðγ  1Þ=2γ, then M2 ≈ 0:397 and M1 ≈ 7:0. Therefore, 9.528 (strong shock approximation) 57.167 (strong shock approximation) 6.0 (strong shock approximation) 0.021 (strong shock assumption)

Comments – The strong-shock approximation leads to the stagnation pressure having the largest deviation among all fluid properties, while the static pressure shows the best approximation. Overall, the approximation yields much better results for M1 ≈ 7:0 than forM1 ≈ 5:0. Other criteria, such as p2 =p1 , can be used, but the ratio pt2 =pt1 is the most stringent, consistent with the βdependence shown in Eqns. (4.19c) through □ (4.19f). For small values of β (weak shock approximation), let M12 ¼ 1 þ , where  ≪ 1 and after substituting this expression into Eqn. (4.15), M2 becomes (John and Keith, 2006) M22 ¼ 1   ¼ 2  M12

(4.20a)

The derivation is straightforward and is left as a problem at the end of the chapter. For the weak normal shock wave, the fluid properties are found by substituting 2 M1 ¼ 1 þ  into Eqn. (4.16):  T2 2γðγ  1Þ  2 γ1 β ≈1 þ M1  1 ¼ 1 þ 2 γþ1 T1 ðγ þ 1Þ

(4.20b)

 p2 2γ  2 M1  1 ¼ 1 þ β ≈1 þ γþ1 p1

(4.20c)

 V2 2  2 β M1  1 ¼ 1  ≈1 γþ1 γ V1

(4.20d)

 ρ2 V1 2  2 β M1  1 ¼ 1 þ ≈ ¼1þ γþ1 γ ρ1 V2

(4.20e)

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4.5 Strong and Weak Normal Shock Waves

 2 3 γ þ 1 3 s2  s1 2γ ≈ M1  1 ¼ β 2 12γ2 R 3ð γ þ 1Þ

   In Eqn. (4.20f), M12  1 was replaced with pp21  1 γþ1 from (4.20c). 2γ

69

(4.20f)

Find the upstream Mach number for a good weak shock approximation.

EXAMPLE 4.5

Solution – For a weak normal shock, assume the criterion to be p2 =p1 ¼ 1:05. Then for γ ¼ 1:4, M12  1 ¼  ¼ 0:043, for which M1 ≈ 1:02.  p2 2γ  2 M1  1 ¼ 1:050 ¼1þ 1:050 ðgivenÞ γ þ 1

p1  T2 γ1  2 ≈1 þ 2 1:013 ðnormal shock relationÞ M1  1 ¼ 1:014 γþ1 T1  ρ2 V1 2  2 M1  1 ¼ 1:037 ≈ ¼1þ 1:033ðnormal shock relationÞ γþ1 ρ1 V2 3 2γ  2 M1  1 2 pt2 ¼ eðs2 s1 Þ=R ≈ e3ðγ þ 1Þ ¼ 0:9999 ≈ 1 ≈ 1 ðnormal shock relationÞ pt1   If p2 =p1 ¼ 1:1. Then for γ ¼ 1:4, M1 ≈ 1:04 and M12  1 ¼  ¼ 0:086.  p2 2γ  2 M1  1 ¼ 1:10 ¼1þ 1:095 ðgivenÞ γ þ 1

p1  T2 γ1  2 ≈1 þ 2 1:026 ðnormal shock relationÞ M1  1 ¼ 1:029 γþ1 T1  ρ2 V1 2  2 M1  1 ¼ 1:072 ≈ ¼1þ 1:067 ðnormal shock relationÞ γþ1 ρ1 V2 3 2γ  2 M1  1 2 pt2 ¼ eðs2 s1 Þ=R ≈ e3ðγ þ 1Þ ¼ 0:9998 ≈ 1 0:9999 ðnormal shock relationÞ pt1   If p2 =p1 ¼ 1:2. Then for γ ¼ 1:4, M1 ≈ 1:08 and M12  1 ¼  ¼ 0:166.  p2 2γ  2 M1  1 ¼ 1:19 ¼1þ γ þ 1

p1  T2 γ1  2 ≈1 þ 2 M1  1 ¼ 1:055 γþ1 T1  ρ2 V1 2  2 M1  1 ¼ 1:139 ≈ ¼1þ γþ1 ρ1 V2 3 2γ  2 M1  1 2 pt2 ¼ eðs2 s1 Þ=R ≈ e3ðγ þ 1Þ ¼ 0:9998 pt1

1:1941 ðgivenÞ 1:0522 ðnormal shock relationÞ 1:1349 ðnormal shock relationÞ 0:9994 ðnormal shock relationÞ

Comments – The accuracy of the weak normal shock approximation appears to be more limited by the density, as values of other fluid properties deviate less from the □ predictions using the full normal shock relation. Downloaded from https:/www.cambridge.org/core. Columbia University Libraries, on 24 Jun 2017 at 04:57:25, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/9781316014288.005

70

4 One-Dimensional Flow, Shock, and Expansion Waves (b)

(a) θ M>1

θ M>1

δ

δ δ θ

(c)

(d)

P∞

P∞

M1

M>1

(e) P∞ M>1

M>1

Figure 4.10 Attached oblique shock waves formed (a) over a two-dimensional inclined surface, (b) over a two-dimensional wedge, and (c) downstream of a supersonic nozzle exit when the exit flow pressure is less than that in the ambient; (d) Normal shock forms at the supersonic nozzle exit plane when the ambient pressure is sufficiently higher than that of the exit flow (pe < p∞ ¼ pb ); (e) when exit flow pressure matches or is greater than the ambient pressure, no shock wave forms and the nozzle (of the flow). For pe ¼ p∞ ¼ pb the nozzle flow is perfectly expanded and for pe > p∞ ¼ pb , under-expanded.

V2

θ

V2t δ

Figure 4.11 Velocity components and inclination angles across an oblique shock wave.

V1n V2n V1t

V1

4.6 Oblique Shock Waves Oblique shock waves are commonly seen in a supersonic flow deflected by a surface that is at an inclined angle to the flow. Consider the planar oblique shock wave formed by a two-dimensional wedge and by a two-dimensional corner with a deflection (or turning) angle δ, which cause the shock wave to be inclined at an

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4.6 Oblique Shock Waves

71

angle of θ to the flow, as shown in Figs. 4.10a and 4.10b, respectively. Such a flow can also be constructed by superimposing a uniform velocity component parallel to the normal shock (V1t ¼ V2t ¼ Vt ) as shown in Fig. 4.11 (the validity of this superimposition is shown by Eqn. (4.21d)). Because the magnitude of the normal velocity component decreases across the shock wave (V1n > V2n ) the resultant velocity downstream of the shock is turned through an angle δ (here δ > 0 for turning into the flow), which varies with the magnitude of the superimposed tangential component. In this case, the surrounding fluid, being at a higher pressure, pushes toward the exit flow with a tangential velocity component. Here δ > 0 if the turning is made into the flow and thus causes compression. The case for δ < 0 will be seen in PrandtlMeyer expansion, the subject of Section 4.9. An oblique shock can thus be formed without a physical wedge, an example of which is the oblique shock wave formed at the exit of a nozzle, as shown in Fig. 4.10c if the pressure of the surrounding is sufficiently high but still lower than that required for a normal shock wave to form. When the surrounding pressure (p∞ , or interchangeably, the back pressure, pb ) is sufficiently high, a normal shock wave will form at the exit plane of the nozzle (Fig. 4.10d), and even within the diverging section of it. Only with pb ≤ pe would there be no shock wave of any type formed for the nozzle flow under this condition; the exit flow is supersonic (Fig. 4.10e). The location of the normal shock wave in the nozzle and the conditions for formation of the normal and oblique shocks will be discussed in Chapter 5. Consider a uniform flow having negligible frictional and body forces passing across an oblique shock. By placing the control volume and control surface shown in Fig. 4.11, the governing equations for the normal component velocity are the same as those for the normal shock wave: ρ1 V1n ¼ ρ2 V2n

(4.21a)

2 2 p1 þ ρ1 V1n ¼ p2 þ ρ1 V2n

(4.21b)

1 2 1 2 ¼ h2 þ V2n h1 þ V1n 2 2

(4.21c)

The momentum equation for the tangential velocity component becomes ð
 ~ ¼0 ~ ⋅ dA Vt ρV CS

Therefore, V1t ðρ1 V1n Þ  V2t ðρ2 V2n Þ ¼ 0 Because ρ1 V1n ¼ ρ2 V2n , this expression becomes V1t ¼ V2t

(4.21d)

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4 One-Dimensional Flow, Shock, and Expansion Waves

Due to the similarity Eqns. (4.10) through (4.12) bear with Eqn. (4.21), one finds that the shock relations from Eqn. (4.15) to (4.18) for the normal shock are applicable to the oblique shock for the normal velocity component. This means that, M2n and all the ratios of temperature, density, and pressure can be determined using the normal shock relations with M1 replaced by M1n in these relations; similarly, Table B.2 can be used as long as M1n is known. Therefore the oblique shock relations are 2 M2n ¼

2 2 M1n þ γ1 2γ 2 γ1 M1n

1

" #γ=ðγ1Þ " #1=ðγ1Þ γþ1 2 pt2 1 ðs2 s1 Þ=R 2 M1n ¼e ¼ 2γ γ1 2 2 pt1 1 þ γ1 2 M1n γþ1 M1n  γþ1 2 p2 2γM1n γ1  ¼ p1 γþ1 γþ1

(4.22a)

(4.22b)

(4.22c)



γ1 2γ 2 2 1 þ M M  1 1n 1n 2 γ1 T2 h i ¼ ðγþ1Þ2 T1 2 2ðγ1Þ M1n

(4.22d)

pffiffiffiffiffiffiffiffiffiffiffi 2 ðγ þ 1ÞM1n ρ2 V1n M1n γRT1 pffiffiffiffiffiffiffiffiffiffiffi ¼ ¼ ¼ 2 þ2 ρ1 V2n M2n γRT2 ðγ  1ÞM1n

(4.22e)

Tt2 ¼1 Tt1

(4.22f)

The absolute Mach number M2 is the vector sum of the normal and tangential components. By using V1t ¼ V2t , pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi M1t γRT1 ¼ M2t γRT2 M2t is determined as rffiffiffiffiffiffi T1 M2t ¼ M1t T2

(4.23)

Therefore, M2 ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ M2 M2n 2t

(4.24)

It is necessary to first determine M1n (and knowing M1n , T2 =T1 can be found) and M1t . From the velocity diagrams shown in Fig. 4.6,

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4.6 Oblique Shock Waves

73

M1t ¼ M1 cosθ; V1t ¼ V1 cosθ

(4.25a)

M1n ¼ M1 sinθ; V1n ¼ V1 sinθ

(4.25b)

M2t ¼ M2 cosðθ  δÞ; V2t ¼ V2 cosðθ  δÞ

(4.25c)

M2n ¼ M2 sinðθ  δÞ; V2n ¼ V2 sinðθ  δÞ

(4.25d)

It is interesting to note that the downstream Mach number of the oblique shock does not have to be less than unity, as required for the normal shock. However, its normal component must be subsonic, i.e., M2 sinðθ  δÞ < 1, although it is possible that M2 > 1. The value of M2 depends on the value of ðθ  δÞ, which in turn depends on the combination of M1 and δ, as shown in Example 4.6 below. Air at 30°C and 200 kPa travels at Mach 4.0. (a) Find the speed, temperature, pressure, shock strength, entropy increase, and pressure loss after the air passing through a normal shock. (b) For comparison, if the air flow experiences an oblique shock at an angle θ ¼ 30° , find the same parameters downstream of this oblique shock. (c) In case of oblique shock, also find the angle through which the air flow is deflected. Assume for air cp ¼ 1:004 kJ=kg ⋅ K.

EXAMPLE 4.6

Solution – (a) For the normal shock, M22

¼

2 M12 þ γ1 2γ 2 γ1 M1

2 16 þ 0:4 ¼ 0:1892; M2 ¼ 0:4350 ¼ 2:8  1 0:4  16  1

 
   γ1 2γ 2 2 1 þ M M  1 1 þ 0:4  16 21:4 1 1 2 γ1 T2 2 h 0:4  16  1 i h i ¼ ¼ ¼ 4:0469; ðγþ1Þ2 2:42 T1 2 2ð1:41Þ  16 2ðγ1Þ M1


T2 ¼ 1; 226:2 K V2 ¼ M2

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi γRT2 ¼ 0:4350  1:4  287  1; 226:2 ¼ 305:3 m=s

p2 2γM12 γ  1 2  1:4  16 1:4  1 ¼  ¼ 18:50; p2 ¼ 3; 700 kPa  ¼ 1:4 þ 1 1:4 þ 1 p1 γ þ 1 γ þ 1 Δp p2 ¼  1 ¼ 17:50 ðshock strengthÞ p1 p1

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74

4 One-Dimensional Flow, Shock, and Expansion Waves

T2 p2  Rln T1 p1 ¼ ð1:004 kJ=kg ⋅ KÞ lnð4:0469Þ  ð0:287 kJ=kg ⋅ K Þlnð18:5Þ ¼ 0:5661 kJ=kg ⋅ K s2  s1 ¼ cp ln

31=ðγ1Þ 3γ=ðγ1Þ 2 γþ1 2 M1 6 7 pt2 6 1 7 2 6 7 ¼4 5 4 2γ 5 γ1 2 γ  1 pt1 2 1þ M1 M1  2 γþ1 γþ1 31=0:4 2 31:4=0:4 2 1:4 þ 1  16 6 1 1:4  17 6 7 2 6 7  ¼4 ¼ 0:1388 5 42  1:4 1:4  1 1:4 þ 15 1þ  16  16 2 1:4 þ 1 2

pt1 ¼ p1 Δpt ¼



1:4=0:4

γ  1 2 γ=ðγ1Þ 1:4  1 M1  16 1þ ¼ 1þ ¼ 151:52 2 2

pt2 pt1 1 p1 ¼ ð1  0:1388Þ  151:52  200 kPa ¼ 26:098 MPa pt1 p1

(b) For the oblique shock, the first parameter to know is M1n . M1n ¼ M1 sinθ ¼ 4 sin30° ¼ 2

2 M2n ¼

2 2 M1n þ γ1

2 4 þ 0:4 ¼ 0:3333; M2n ¼ 0:5774 ¼ 2γ 2:8 2 0:4  4  1 γ1 M1n  1



γ1 2γ 2 2 1 þ M M  1 1n 1n 2 γ1 T2 h i ¼ ðγþ1Þ2 T1 2 2ðγ1Þ M1n  21:4  1 þ 0:4 41 2 h 4 0:4 i ¼ ¼ 1:6875; T2 ¼ 511:3 K 2:42  4 2ð1:41Þ

rffiffiffiffiffiffi T1 M2t ¼ M1t ; T2

rffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi rffiffiffiffiffiffi T1 T1 1 ° ° ¼ 2:6667 M2t ¼ M1t ¼ M1 cos30 ¼ 4 cos30 1:6875 T2 T2 M2 ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ M2 ¼ M2n ð0:5774Þ2 þ ð2:6667Þ2 ¼ 2:7285 2t

It is noted that in this case the contribution of M2n to M2 is relatively small.

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4.6 Oblique Shock Waves

V2 ¼ M2

75

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi γRT2 ¼ 2:7285  1:4  287  511:3 ¼ 1; 236:7 m=s

2 p2 2γM1n γ  1 2  1:4  4 1:4  1 ¼  ¼ 4:50; p2 ¼ 900 kPa  ¼ 1:4 þ 1 1:4 þ 1 p1 γþ1 γþ1

Δp p2 ¼  1 ¼ 3:50 p1 p1 (This shock strength is five times smaller than that across a normal shock with the same upstream Mach number.) T2 p2  Rln ¼ T1 p1 ð1:004 kJ=kg ⋅ KÞ lnð1:6875Þ  ð0:287 kJ=kg ⋅ K Þlnð4:5Þ ¼ 0:0936 kJ=kg ⋅ K

s2  s1 ¼ cp ln

31=ðγ1Þ 3γ=ðγ1Þ 2 γþ1 2 M1n 6 7 pt2 6 1 7 2 6 7 ¼4 5 4 γ1 2 2γ γ  15 pt1 2 M1n M  1þ 2 γ þ 1 1n γ þ 1 31=0:4 2 31:4=0:4 2 1:4 þ 1 4 6 1 1:4  17 6 7 2 6 7  ¼4 ¼ 0:7209 5 4 1:4  1 2  1:4 1:4 þ 15 4 4 1þ 2 1:4 þ 1 2

Δpt ¼



pt2 pt1 1 p1 ¼ ð1  0:7209Þ  151:52  200 kPa ¼ 8:458 MPa pt1 p1

(c) The flow deflection angle can be found using either Eqn. (4.24c) or Eqn. (4.24d).   M2n 0:5774 ¼ sinðθ  δÞ ¼ sin 30°  δ ; δ ¼ 17:78° ¼ 2:7285 M2 Comments – As discussed in the text, when the back pressure is insufficient to cause a normal shock (900 kPa < 3,700 kPa needed for a normal shock), an oblique shock may result. Both the pressure and temperature rises across an oblique shock are smaller than those across a normal shock, as an oblique shock is a weaker compression wave than a normal shock; it also causes a smaller entropy increase and thus a smaller pressure loss. The flow deflection angle can also be a result of a two-dimensional wedge (or surface) at an angle of 17:78° measured from the flow direction.

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4 One-Dimensional Flow, Shock, and Expansion Waves

The downstream properties can also be found after M2 is calculated. As an illustra
γ=ðγ1Þ  1:4=0:4 2 M ¼ 1 þ 1:41 ¼ 24:32 tion, for M2 ¼ 2:7285, ppt22 ¼ 1 þ γ1 2 2 2  7:4445Þ p2 ¼

pt2 0:7209pt1 0:7209  151:52  200 kPa ¼ 898:1 kPa ≈ 900 kPa ¼ ¼ 24:32 24:32 24:32

□

Example 4.6 illustrates that the flow deflection angle (δ) and the downstream Mach number M2 can be calculated by knowing the incoming flow Mach number (M1 ) and the inclination of the shock wave angle (i.e., θ). It is then desirable to obtain an expression of θ in terms of δ and M1 . The ratio of normal velocity components (Eqn. [4.22e]) is: 2 ðγ þ 1ÞM1n ðγ þ 1ÞM12 sin2 θ V1n V1t tanθ tanθ ρ ¼ 2¼ ¼ ¼ ¼ 2 V2n V2t tanðθ  δÞ tanðθ  δÞ ρ1 ðγ  1ÞM1n þ 2 ðγ  1ÞM1 sin2 θ þ 2

(4.26) After some trigonometric manipulations, tanδ ¼ 2cotθ

M12 sin2 θ  1 2 M1 ðγ þ cos2θÞ þ

2

(4.27)

For δ ¼ 0, M12 sin2 θ ¼ 1 and this occurs in two scenarios: (i) for θ ¼ π=2 (i.e., the normal shock, for which no flow redirection occurs immediately downstream of the shock) and (ii) θ ¼ sin1 ð1=M1 Þ; in this case θ is also the Mach wave angle μ as expressed by Eqn. (3.12). Therefore for δ > 0, θ is expected to fall in the range given by sin1

1 π ≤θ≤ M1 2

(4.28)

The relation between θ and δ, Eqn. (4.27), is graphically shown in Fig. 4.12 for γ ¼ 1:4. A few observations and comments can be made: (1) As shown in Fig. 4.12, for a given pair of M1 and δ, there are in general two possible oblique shock wave angles – one for a weak shock (smaller θ value associated with smaller M1n ) and the other for a stronger shock (largerθ value associated with larger M1n ). Usually, the weak oblique shock appears and, in the case of a two-dimensional wedge as the flow turning device, the shock wave attaches to the vertex. For a given M1 , if the deflection angle δ is too large, there is no solution as represented in Fig. 4.11. For example, for M1 ¼ 3, there is no attached shock solution (i.e., no value of θ exists) for δ greater than approximately 34°. The flow is forced to adjust to it through a detached normal shock instead of an oblique shock, as illustrated in Fig. 4.13.

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4.6 Oblique Shock Waves

77

45°

40°

δ = δs (M2 > 1)

Deflection angle (δ)

30° M1 = ∞ 10

25°

M2 > 1

δ = δmax

35°

M2 < 1

5 20°

4 3

15° 2 10°

1.6 1.4

5° 1.2 0° 0°

10°

20°

30°

40° 50° Wave angle (θ)

60°

70°

80°

90°

Figure 4.12 Oblique shock solutions, expressed by Eqn. (4.27) for γ ¼ 1:4 (adapted from Liepmann, H. W., and Roshko, A., Elements of Gasdynamics, John Wiley and Sons, New York, 1957).

(2) The supersonic nozzle exit flow discussed earlier may enter the surroundings without encountering physical deflecting object (or obstruction). Depending on the pressure of the surroundings (or rather the pressure ratio of the surroundings to the exit gas), an oblique shock may form with the flow turning angle and the shock wave angle given by Eqn. (4.27). In the case of δ ¼ 0, either a normal shock (θ ¼ π=2, Fig. 4.10d) due to sufficiently high pressure ratio p2 =p1 given by Eqn. (4.16a) or no shock of either type (normal and oblique, Fig. 4.10e) is formed. In the latter case, the angle θ ¼ sin1 ð1=M1 Þ is the Mach cone angle (or the Mach wave angle) described in Chapter 3 implying that the pressure gradient is zero. In this case of a nozzle exit flow, the flow is said to be perfectly expanded (to the ambient pressure, pb , as shown in Fig. 4.10e). In contrast, the flow that forms either a normal or an oblique shock at the exit is said to be overexpanded because its pressure falls below that of the ambient (i.e., pe < pb ) and, due to its supersonic speed, it adjusts to the ambient condition through a shock wave.

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4 One-Dimensional Flow, Shock, and Expansion Waves

(3) In this θ  δ plot, two contours for the particular flow deflection angles are also shown: (i) δmax for causing the maximum shock wave angle θmax and (ii) δs for causing sonic downstream flow (ðM2 ¼ 1Þ. The value of θmax represents the maximum shock wave angle with which the oblique shock remains attached. They were determined from the following expressions (Ferri, 2005): ( 

 ) 1 γþ1 2 γ  1 2 γ þ 1 4 1=2 2 M1  1 þ ðγ þ 1Þ 1 þ M1 þ M1 (4.29) sin θmax ¼ 4 2 16 γM12

1 sin θs ¼ γM12 2

(



 ) γþ1 2 3γ γ þ 9 3  γ 2 γ þ 1 4 1=2 M1  þ ð γ þ 1Þ  M1 þ M1 4 4 16 8 16 (4.30)

For a given M1 there are corresponding values of δ that would cause θmax and M2 ¼ 1. As an example, for M1 ¼ 3:0, δs ≈ δmax ≈ 34° with corresponding θmax ≈ 65° . For M1 ¼ 1:4 and δs ¼ 8:5° (θs ¼ 63:5° ) and δmax ≈ 9:25° and θmax ≈ 67:5° . As a consequence, for M1 ¼ 1:4 the shock wave becomes detached for δ > 9:25° . Therefore for M1 ¼ 1:4 and 9:25° > δ > 9° the oblique shock is attached with M2 < 1. (4) There exist subsonic values of M2 for a narrow range of δ for a given M1 . For M1 ¼ 1:6, for example, M2 < 1 for δ ≈ 14:3°  14:7° , as shown by the solid line segment between the δ ¼ δs and δ ¼ δmax curve. For most values of δ, the weak solution results in M2 > 1. Strong solutions require the shock to detach from the wedge and M2 < 1. Figure 4.12 can be used to illustrate how the shock wave angle may evolve as M1 increases. Consider a two-dimensional wedge with a half wedge angle δ ¼ 15° symmetrically aligned with the flow. As M1 increases from subsonic condition to approximately M1 ¼ 1:6, there is no oblique solution – instead there is either no shock (subsonic flow) or detached normal shock as the horizontal line of δ ¼ 15° and the solution curves for these Mach numbers do not intersect. As M1 increases to approximately 1.62, the oblique shock forms with a wave angle θ ≈ 63° . When M1 reaches 2, 3, 4, 5, and so on, and toward infinity, the oblique shock wave angle θ ≈ 45° ; 32° ; 27° ; 23° and toward 18° , respectively. The Mach number that divides the regimes of attached and detached shocks is called the detachment Mach number. The result of Fig 4.12 indicates that this number is dependent on the deflection angle. For δ ¼ 15° , it is 1.62, as discussed, and for δ ¼ 10° , it is approximately equal to 1.43. Oblique shock solutions similar to Fig. 4.12 with higher resolution are found in Appendix E. The results of Fig. 4.12 suggest that for a given δ, the difference between θ and δ decreases as M1 increases. High Mach number flows are thus more able to “press” onto the wedge. For M1 ! ∞, as shown in Fig. 4.12, θ is larger than δ by a decreasingly smaller fraction. For example for δ ¼ 10° , θ ≈ 12° for M1 ! ∞ and, for

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4.6 Oblique Shock Waves

(a)

79

1 θ = μ = Sin–1( M ) (far-field Mach wave) 1

Sonic bubble

bow shock M2 > 1 M2 < 1 δ > δmax

M1 > 1 a

δ > δmax

(b)

1 θ = μ = Sin–1( M

1

) (far-field Mach wave)

c bow shock

b

Sonic bubble

M2 < 1 M1 > 1

M2 > 1

a

c

(c)

the slip line o

e a

b

Figure 4.13 (a) Schematic of (a) the detached shock wave system on a wedge with an afterbody, (b) the detached shock wave system on a blunt-nose body, and (c) merging of oblique shocks to form a stronger shock over a concave corner. Note in (a) and (b) that the far-field shock wave becomes the Mach wave, while in (c) the entropy change through the three oblique shocks is different from that through the merged shock, leading to the formation of the slip line.

comparison, θ ≈ 51° for M1 ¼ 1:6. As will be seen in Chapter 12 on hypersonic flows, for Eqn. (4.33c), θ ! ðγ þ 1Þθ=2 as M1 ! ∞. Figure 4.13a qualitatively illustrates a detached shock wave system on a wedge with an afterbody. At point a along the shock, the wave angle θ is equal to π=2. In this

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4 One-Dimensional Flow, Shock, and Expansion Waves

region, the shock wave is a normal shock and the downstream Mach number M2 < 1. Moving away from point a, the shock wave becomes weaker and beyond point b (where M2 ¼ 1), the downstream flow is supersonic. Between points a and b, where the flow is subsonic, the effect of the shoulder can be felt by the flow, helping to explain the curvature of the shock wave. Moving further away from point b, the shock becomes so weak that its inclination (θ) asymptotes to the Mach angle, as if not affected by the wedge or the afterbody. As an example, for M1 ¼ 2:0, the inclination
 is 30° ¼ μ ¼ sin1 ½1=M1  , the same as the Mach (or wave) angle for δ ¼ 0° as can be found on Fig. 4.12. Therefore, conditions along a detached shock wave correspond to the entire range of the curve for a given M1 . Different values of M1 not only give different asymptotic values of θð¼ μÞ, but also different values of θmax ( ≈ 67:5° for M1 ¼ 1.4 and ≈ 64:4° for M1 ¼ 2:0). Referring to Fig. 4.13, one can expect that the shape of the detached shock will depend on M1 for a given geometry. As can be seen in Fig. 4.11, one would also expect that flows having smaller M1 detach at smaller δ. It is also natural to expect it to depend on the geometry for a given M1 . Figure 4.13b depicts the detached shock wave system on a blunt-nose body, which can be thought of as having δ > δmax at the nose. The wave near point a has a larger segment for θ ¼ π=2 and M2 < 1 than the corresponding point a in Fig. 4.13a. Because of the curvature, the shock strength gradually diminishes along a detached shock away from point a in both Figs. 4.12a and 4.12b. The resultant gradient of entropy would cause generation of vorticity, even if the incoming flow is uniform and without vorticity – this will be shown in Section 8.7. The associated entropy rise also decreases along the shock wave in the direction toward the far field. Oblique shocks may merge to form a stronger shock wave, as shown in Fig. 4.13c. One expects the entropy increase behind segment oc to be larger than that behind segment oe. A dividing line might be identified (the “slip” line as shown in Fig. 4.13c), across which entropies do no match. Along a gradually curved shock wave, such as those shown in Figs. 4.12a and 4.12b, the slip line region is more diffuse, and thus forms an entropy layer. For attached shocks, once the corresponding values of θ and δ are found, the downstream Mach number can be calculated by combining Eqns. (4.22a), (4.25b), and (4.25d), so that M2 is a function of M1 and δ M22 sin 2 ðθ  δÞ ¼

2 M12 sin 2 θ þ γ1 2γ 2 γ1 M1

sin 2 θ  1

(4.31)

where it is understood that θ is also a function of M1 and δ. By referring to Fig. 4.13 and Eqn. (4.22), all shock waves, including bow shocks, can be treated as normal shocks, as long as the local normal component of the Mach number of the approaching flow is used (Thompson, 1972). For this reason, the solution process in Example 4.6 can be followed along with using the table in Appendix D.

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4.6 Oblique Shock Waves

81

For the same oblique shock wave in Example 4.6, use Fig. 4.12 to find the downstream Mach number. Also compare the shock strengths for M1 ¼ 4 and 2 for the same wedge angle.

EXAMPLE 4.7

Solution – For M1 ¼ 4 and θ ¼ 30° , Fig. 4.12 indicates that δ ≈ 17:75° . Eqn. (4.31) becomes 2 M12 sin 2 30° þ γ1   M22 sin 2 30°  17:5° ¼ 2γ 2 2 ° γ1 M1 sin 30  1

M2 ≈ 2:67 This value is slightly different from that calculated in Example 4.4, due to the different values of δ. For M1 ¼ 2 and δ ≈ 17:75° , Fig. 4.7 gives θ ≈ 48:5° . Thus M1n ¼ M1 sinθ ¼ 2  sin 48:5° ≈ 1:50. Form the normal shock table, pp21 ¼ 2:458 and β ¼ pp21  1 ¼ 1:458. M1 ¼ 4, M1n ¼ M1 sinθ ¼ 4  sin30° ¼ 2. Therefore, pp21 ¼ 4:50 and β ¼ pp21  1 ¼ 3:50. Comments – As M1 is increased for a given wedge angle, the shock strength increases even though the shock wave angle decreases. This is because M1n ¼ M1 sinθ still increases as long as the shock remains attached, further illustrating that higher Mach number flows are more able to “press” on the wedge, with higher pressure □ on the surface and more inclined shock wave. A useful alternative to Eqn. (4.27) for relating δ and θ can be derived by rearranging Eqn. (4.26) so that 1 M12 sin2 θ

¼

γ þ 1 tanðθ  δÞ γ  1  2 tanθ 2

(4.32)

Applying necessary trigonometric identities, this expression is reduced to M12 sin2 θ  1 ¼

γ þ 1 2 sinθ sinδ M1 2 cosðθ  δÞ

(4.33a)

The usefulness of Eqn. (4.33a) can be seen in large-Mach number flows, where small turning angles are desirable, i.e., δ ≪ 1, so that sin δ ≈ δ and cosðθ  δÞ ≈ cosθ. In these flows,

γþ1 2 2 2 M1 tanθ ⋅ δ M1 sin θ  1 ¼ (4.33b) 2

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4 One-Dimensional Flow, Shock, and Expansion Waves

For extremely large Mach numbers (M1 ! ∞), results from Fig. 4.12 suggest θ ≪ 1. Using small-angle approximation for Eqn. (4.33b), sinθ ≈ tanθ ≈ θ, and M12 sin2 θ ≫ 1, one finds θ¼

γþ1 δ 2

(4.33c)

Equation (4.33c) appears to be an excellent approximation even for moderately large deflection angles. For and M1 ! ∞ and M1 ¼ 10, the θ  δ plot preserves the slope of ðγ þ 1Þ=2 ≈ 1:2 for δ up to and beyond 25° (i.e., δ ¼ 0:436). The flow with M1 ! ∞ is called the hypersonic flow. More details related to hypersonic flows are presented in Chapter 12. For M1 ! ∞, find the attached oblique shock solution for the deflection angle, shock wave angle, downstream Mach number. Consider M1 ¼ 10 as approaching ∞. Find ratios of pressure and temperature across the shock wave for γ ¼ 1:4. EXAMPLE 4.8

Solution – The maximum shock wave angle θmax can be directly determined without knowing δ, according to Eqn. (4.29): ( 

 ) 1 γþ1 2 γ  1 2 γ þ 1 4 1=2 2 M 1  1 þ ð γ þ 1Þ 1 þ M1 þ M1 sin θmax ¼ lim M1 !∞ γM2 4 2 16 1 ¼

γþ1 2γ θmax

sffiffiffiffiffiffiffiffiffiffiffi γþ1 ¼ sin 2γ 1

Then tanδ max ¼ lim 2cotθ M1 !∞

¼

M12 sin2 θ  1 cosθmax sin2 θmax ¼ 2 sinθmax γ þ cos2θmax M12 ðγ þ cos 2θÞ þ 2

sin2θmax γ þ cos2θmax

For γ ¼ 1:4, θmax ¼ 67:792° δmax ¼ 45:579° These values are consistent with those than can be read from Fig. 4.12. M2 sin 2 θ þ 2 M22 sin 2 ðθ  δÞ ¼ 2γ1 2 2 γ1 ¼ γ1 2γ in the limit of M1 ! ∞. γ1M1

sin θ1

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4.7 Weak Oblique Shock

83

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi γ1 0:4 ¼ M2 ¼   ¼ 0:9998 ≈ 1 2 2γ sin 2 ðθ  δÞ 2  1:4  sin 67:792°  45:579° For M1 ¼ 10, M1n

sffiffiffiffiffiffiffiffiffiffiffi γþ1 ¼ 9:26 ¼ M1 sinθ ≈ M1 2γ

2 p2 2γM1n γ1 ≈ M12 ¼ 100ð ≈ 100 from Appendix DÞ  ¼ p1 γþ1 γþ1

2 T2 2γðγ  1ÞM1n ð γ  1Þ 2 M ≈ 16:7ð ≈ 17:5 from Appendix DÞ ¼ ¼ 2 ð γ þ 1Þ 1 T1 ðγ þ 1Þ

Comments – This near-unity value of the downstream Mach number is to be expected, as Eqns. (4.29) and (4.30) for θmax and θs (shock wave angle for M2 ¼ 1) are the same as M1 ! ∞. This result can be seen in Fig. 4.12. From Appendix D, the pressure and □ temperature ratios are approximately 100 and 17.5, respectively.

4.7 Weak Oblique Shock As discussed in Section 4.6, shock waves formed with moderate values of M1 are prone to detachment. Small turning (or deflection) angles are desirable to avoid detachment for this range of Mach numbers. For the limiting case where δ ! 0, either the right-hand side of Eqn. (4.33b) is finite and tanθ ! ∞ (i.e., θ ¼ π=2) or the left-hand side approaches zero (i.e., θ ¼ sin1 ð1=M1 Þ ¼ μ, the Mach angle), for attached oblique shock waves θ ≤ θmax (as shown in Fig. 4.12). As a consequence, tanθ has to be finite and therefore 1 tan θ ¼ tan μ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M12  1 Equation (4.33b) now becomes 0

1

Bγ þ 1 C qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA ⋅ δ M12 sin2 θ  1 ¼ @ 2 2 M1  1 M12

(4.34)

The shock strength depends on M1n ¼ M1 sinθ, and Eqn. (4.16e) is rewritten for the weak oblique shock (with small δ) as

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4 One-Dimensional Flow, Shock, and Expansion Waves

0 β¼

1

 B p 2  p1 2γ  2 C M1 sin2 θ  1 ¼ @qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA ⋅ δ ¼ γþ1 p1 M2  1 γM12

(4.35)

1

Therefore for weak oblique shocks, the shock strength is proportional to the turning angle, δ.

4.8 Compression in Supersonic Flow by Turning Similar to the weak normal shock wave, the weak oblique shock causes an entropy increase proportional to the third power of the shock strength. Eqn. (4.22f) for the normal shock can be written in terms of the normal component of the Mach number M1n :  2 3 γ þ 1 3 3 s2  s1 2γ ≈ M1n  1 ¼ β ∝δ 2 12γ2 R 3ð γ þ 1Þ

(4.36)

For small values of δ, the entropy increase becomes vanishingly smaller and so does the loss in stagnation pressure, as   pt1  pt2 ¼ 1  eðs2 s1 Þ=R ¼ O δ3 pt1

(4.37)

where binomial expansion was used for the exponential term for small ðs2  s1 Þ=R. This result is similar to those shown in Example 4.5 for weak normal shocks, where the loss in the stagnation pressure is vanishingly negligible with a small supersonic upstream Mach number. For weak oblique shock waves, the entropy increase and pressure loss are now related to the turning angle, as shown in Eqns. (4.36) and (4.37). It is thus desirable to achieve a turning angle (i.e., compression) through a series of weak oblique shocks using smaller turning angles than by one single oblique shock through a finite turning angle. Figure 4.14a shows an oblique shock formed at a small finite turning angle δ. To further minimize entropy increase due to turning, let nΔδ ¼ δ, where n is the number of much smaller equal turning angles Δδ that comprise the finite turning angle δ, as shown in Figs. 4.14a and 4.14b. Equations (4.36) and (4.37) can be written for each Δδ as ΔβeΔδ ΔseðΔδÞ3 The overall changes after completing the turn are β¼

p2  p1 δ p1 e

(4.38)

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4.8 Compression in Supersonic Flow by Turning

85

(a) 1

2

M1>1 δ

Slip line

(b)

M1>1 Dd Dd

d

Dd

(c) Shock wave

Mach waves

Slip line

Figure 4.14 To turn a supersonic flow through an angle δ (part a) with decreasing amounts entropy increase, the turn can be achieved by gradual turning through a series of n turns, each with Δδ so that nΔδ ¼ δ, allowing the flow to pass a series of weaker shock waves (part b). As n ! ∞, the shock waves become Mach waves. Similar to Fig. 4.12, slip lines form in parts (b) and (c). No slip line exists as the flow experiences uniform entropy increase through the single oblique shock.

d

M1>1

3 Δs s2  s1 ¼ ¼ nðΔδÞ3 enΔδðΔδÞ2 δðΔδÞ2 δ e e R R n2

3

pt1  pt2 δ ¼O 2 pt1 n

(4.39a)

(4.39b)

Recall that for compression (i.e., wedge turning into the flow), δ > 0 and Δδ > 0. Comparing with compression by one single oblique shock, the same degree of compression (β) can be achieved with a much reduced entropy increase and a much reduced pressure loss by a factor of 1=n2 . In the limiting case of Δδ ! 0þ and n ! ∞ the turning is a continuous procedure and the oblique shock waves becomes Mach waves, as illustrated in Fig. 4.14c where some representative Mach lines are shown. In this limiting procedure, the compression is isentropic. The flow between the two successive Mach lines is uniform. However, in the limiting compression process, the region between the representative Mach lines also becomes vanishingly small and the change in flow becomes continuous. The following numerical example illustrates the benefits of such successive turning through smaller angles.

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86

4 One-Dimensional Flow, Shock, and Expansion Waves EXAMPLE 4.9 An air flow at Mach 4.0 and 216 K is being turned by (1) a single turning angle of 10° and (2) two successive turning angles, both of 5°. Assume cp ¼ 1:0 kJ=kg ⋅ K. Compare pressure increases, the stagnation pressure change, entropy change, and the final flow speeds of the two scenarios.

1 M1= 4.0

3

2

5º 5º

1 M1= 4

2 10º

angles are not to the scale

Solution – From Fig. 4.12, for M1 ¼ 4:0 and δ ¼ 10° , θ ≈ 23° . Therefore, M1n ¼ M1 sin 23° ≈ 1:56. Therefore, using shock table or Eqns. (4.16) and (4.17d) yields p2 p2  p1 ¼ 2:673 and β ¼ ¼ 1:673 p1 p1 pt2 pt2  pt1 ¼ 0:910 ¼ 9% pt1 pt1 T2 ¼ 1:361 T1 M2n ¼ 0:681 rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi rffiffiffiffiffiffi T1 T1 1 ° ¼ 3:156 M2t ¼ M1t ¼ M1 cos θ1 ¼ 4 cos 23 1:361 T2 T2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi Therefore M2 ¼ 3:229 andV2 ¼ M2 γRT2 ¼ 3:229  1:4  288  1:361  216 ¼ 1; 111:7 m=s Δs ¼ s2  s1 ¼ cp ln

T2 p2  R ln ¼ 1:0  lnð1:361Þ  0:287  lnð2:673Þ T1 p1 ¼ 0:026 kJ=kg ⋅ K

With successive turning, the first turning results in (with M1 ¼ 4:0 and δ1 ¼ Δδ ¼ 5° ) θ1 ≈ 18° and M1n ¼ M1 sin18° ≈ 1:24. Therefore, using shock table or Eqns. (4.16) and (4.17d) again yields p2 ¼ 1:627 p1

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4.8 Compression in Supersonic Flow by Turning

87

pt2 pt2  pt1 ¼ 0:988 or ¼ 1:2% pt1 pt1 T2 ¼ 1:153 T1 M2n ¼ 0:818 rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi rffiffiffiffiffiffi T1 T1 1 ° ¼ 3:543 M2t ¼ M1t ¼ M1 cos θ1 ¼ 4 cos 18 1:153 T2 T2 Therefore M2 ≈ 3:64 Now with δ2 ¼ Δδ ¼ 5° ,θ2 ≈ 19° . Therefore M2n ¼ M2 sin19° ≈ 1:19. Thus p3 ¼ 1:486 p2 pt3 ¼ 0:993 pt2 T3 ¼ 1:120 T2 Mn3 ¼ 0:848 rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi rffiffiffiffiffiffi T1 T1 1 ° ¼ 3:253 Mt3 ¼ Mt2 ¼ M2 cosθ2 ¼ 3:64 cos19 1:120 T2 T2 M3 ¼ 3:360 and V3 ¼ M3

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi γRT3 ¼ 3:36  1:4  288  1:293  216

¼ 1; 127:5m=s pt3 pt3 pt2 pt3  pt1 ¼ ¼ 0:993  0:988 ¼ 0:981 and ≈ 2% pt1 pt2 pt1 pt1 p3 p3 p2 p3  p1 ¼ ¼ 1:486  1:627 ¼ 2:418 and β ¼ ¼ 1:418 p1 p2 p1 p1 T3 T3 T2 ¼ ¼ 1:121  1:153 ¼ 1:293 T1 T2 T1 Δs ¼ s3  s1 ¼ cp ln

T3 p3  R ln ¼ 1:0  lnð1:293Þ  0:287  lnð2:418Þ T1 p1 ¼ 0:004 kJ=kg ⋅ K

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4 One-Dimensional Flow, Shock, and Expansion Waves

Comments – It can be seen that indeed Δse1=n2 . In this example, n = 2 and the values of Δs differ by a factor of 4.5, which is close to n2 ¼ 4. Furthermore, the pressure loss   t2 ¼ 1  eðs2 s1 Þ=R ¼ O δ3 suggests that by halving the given by Eqn. (4.37), pt1pp t1 angle, the pressure loss should decrease by a factor of 8. The first turning by t1 = 1.2% close to being one-eighth (¼ 1=n3 , with n ¼ 2) of loss due results in pt2pp t1 ° to the 10 turning. If p1 is known, both the static and stagnation pressures before and after the shock waves can be calculated. The static pressure contributes to the force acting on a wing surface and its value is therefore necessary when aerodynamic lift and drag are of interest. For the same total turning angle, □ a series of smaller turning angle results in smaller β. A series of oblique shock waves can be used for pressure recovery in supersonic inlets. The following example illustrates such an application. EXAMPLE 4.10 Consider two different two-dimensional supersonic inlets for a free stream Mach number M1 ¼ 4. One operates with a standing normal shock wave at the inlet as the way to slowing down the flow, the other uses the scheme similar to that described in Example 4.9, with wedge-shaped diffuser, which generate two oblique shocks through two 5° turning angles for deceleration, followed by further deceleration through a normal shock. Compare the stagnation pressure losses.

1

M1= 4 2

M2< 1

M1= 4

3

4

2

Solution – For the one single normal shock for M1 ¼ 4, M2 ¼ 0:435 and pt2 =pt1 ¼ 0:1388 (a loss of approximately 86%). For the second case, M3 ¼ 3:36 from Example 4.9. Therefore through the normal shock, M4 ¼ 0:457 and pt4 =pt3 ¼ 0:2404. The pt4 =pt1 ¼ ðpt4 =pt3 Þðpt3 =pt1 Þ ¼ 0:2404  0:981 ¼ 0:236 and the stagnation pressure loss is approximately 76%. Therefore there is an advantage by adopting the second flow deceleration scheme for supersonic diffuser design, even though the final compression may □ still be done through a normal shock. For small but non-vanishing values of Δδ > 0, the shock wave angle θ deviates from the Mach angle μ slightly. Let the deviation be ε, with ε ≪ μ, such that θ ¼μþε

(4.40)

and

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4.8 Compression in Supersonic Flow by Turning

89

sin θ ¼ sinðμ þ εÞ ¼ sin μ cos ε þ sin ε cos μ ≈ sin μ þ ε cos μ (4.41)  2 1=2 . Dividing the above expression through by Since sin μ ¼ 1=M1 , cot μ ¼ M1  1 sinμ yields  1=2 M1 sinθ ≈ 1 þ ε cot μ ¼ 1 þ ε M12  1

(4.42a)

A series expansion for M12 sin2 θ and neglecting terms containing ε2 leads to  1=2 M12 sin2 θ ≈ 1 þ 2ε M12  1 Comparing with Eqn. (4.34) leads to the following expression:

γ þ 1 M12 ε¼ ⋅ Δδ 4 M12  1

(4.43b)

(4.44)

It is also of interest to find the velocity change across the wave. By referring to Fig. 4.11, with V2t ¼ V1t ¼ Vt one arrives at the following: 2 V22 V2n þ V2t2 ðV2n =Vt Þ2 þ 1 tan2 ðθ  δÞ þ 1 cos2 θ ¼ ¼ 2 ¼ ¼ 2 2 2 2 2 cos ðθ  ΔδÞ tan θ þ 1 V1 V1n þ V1t ðV1n =Vt Þ þ 1

This expression can be rewritten as V2 cosθ cosθ cosθ 1 ¼ ≈ ¼ ¼ cosðθ  ΔδÞ cosθ cosΔδ þ sinθ sinΔδ cosθ þ sinθ ⋅ Δδ 1 þ tanθ ⋅ Δδ V1 where small angle approximations, sinΔδ ≈ Δδ and cosΔδ ≈ 1, are used. Equation (4.42a) can be manipulated to provide information for tanθ by assuming ε ≪ 1 (as ε ≪ μ and μ is of the order of unity or smaller), resulting in 1 tanθ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M12  1 Substituting this expression for tanθ back to the expression for V2 =V1 , and noting that for a small value x, 1=ð1 þ xÞ ≈ ð1  xÞ and letting x ¼ tanθ ⋅ Δδ yields V2 Δδ ¼ 1  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi V1 M2  1

(4.45a)

ΔV V2  V1 Δδ ¼ ¼  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi V1 V1 M2  1

(4.45b)

1

1

In the differential expression, Eqn. (4.45b) becomes dV dδ ¼  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi V M2  1

(4.46)

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4 One-Dimensional Flow, Shock, and Expansion Waves

Because compression makes M and μ decrease while making ε increase, according to Eqns. (4.40) and (4.44), Mach wave and shock wave angles shown in Figs. 4.14a and 4.14c become increasingly steeper. At a sufficient distance away from the turning corner, depending how gradual the turn is, these Mach lines converge to form a shock wave, as shown in Fig. 4.14c. Because there is no length scale to characterize the corner region, the flow and fluid parameters in the near-wall region must only be dependent upon the angle through which the flow is turned. There is a Mach wave angle corresponding to each turning angle and thus the flow and fluid properties must be uniform at locations along the Mach lines from the corner. The flow is uniform between two Mach lines. Since a continuous turn creates an infinite number of Mach lines with infinitesimal spacing, it generates a continuous compression process near the turning corner. Derivations leading to Eqns. (4.15) and (4.16) relating the downstream and upstream Mach numbers do not involve the assumption of a shock wave. In fact a supersonic flow may expand and accelerate to be more supersonic, i.e., M2 > M1 > 1 and p2 < p1 as allowed by p2 1 þ γM12 ¼ p1 1 þ γM22

(4.16a)

Because the flow turning associated with weak oblique shocks can be accomplished by small turning angles, the compression is nearly isentropic. Would supersonic Expansion fan (mach waves): φ1–2 = (μ1– μ2) + Δν1–2

(a)

μ=π⁄2

μ1 = sin–1(M1 ) 1

(M=1)

V1

μ2 = sin–1( M1 )

M1 > 1 M

=1

f ν= ν re

=0

1

1

ν2 = ν1 + sin–1( M1 1)–sin–1( M2)

V1

1

δ1 = ν1

= ν1 + δ

2

δ12 = Δν1–2

M2 > M1

(b) (M=1)

V1

ν1 V2 = V1 + Δν1–2

M1 > 1 M

=1

ν= ν re

f

=0

1

δ1 = ν1 2

M2 > M1 δ12 = Δν1–2

Figure 4.15 Schematic of Prandtl-Meyer expansion from region 1 to region 2 through (a) a sharp turning angle and (b) through a smooth but equal turning angle.

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4.9 The Prandtl-Meyer Function

91

expansion be accomplished with dδ < 0 (i.e., deflection is away from the flow and the flow turns around a convex angle)? The case of dδ < 0 (δ2 < δ1 < 0) ε < 0 constitutes the Prandtl-Meyer expansion discussed in the following section.

4.9 The Prandtl-Meyer Function Imagine a sharp (or “centered”) turn away from the flow (dδ < 0 and δ2 < δ1 < 0, as shown in Fig. 4.15a) as consisting of a series of much more gradual turns through much smaller turning angles that make up the same total turning angle (shown in Fig. 4.15b). Due to the small turning angles, the flow near the corner is isentropic as in the case of isentropic compression. As a result, the fluid and flow properties are constant along each of these Mach waves and the flow is considered uniform in the near-corner region between two successive Mach waves. Eqn. (4.46) is applicable for its analysis and is rewritten as dδ ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dV M2  1 V

(4.47a)

where the + and − signs denote turning away from and into the flow, respectively. Integration of this expression for Prandtl-Meyer expansion yields

ð

þδ þ constant ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dV ¼  ðM Þ M2  1 V

(4.47b)

where  ðMÞ is called the Prandtl-Meyer function. It is desirable to express V in terms of M to obtain an explicit expression only in M. The following relations can be used: V ¼ Ma and a2t γRTt γ1 2 M ¼1þ ¼ 2 a2 γRT

(4.47c)

where the subscript t once again denotes the stagnation state. For isentropic expansion, Tt ¼ constant. It follows that dV dM da ¼ þ V M a

(4.47d)

Because Tt = constant (i.e., a2t ¼ γRT t ). After some algebraic manipulations, da ðγ  1ÞMdM  ¼
a 2 1 þ γ1 M2 2

and

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4 One-Dimensional Flow, Shock, and Expansion Waves

dV dM 1 ¼ V M 1 þ γ1 M2 2

! (4.47e)

By substituting Eqn. (4.47d) into Eqn. (4.47a) the Prandtl-Meyer function becomes ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M2  1 dM γþ1 γ1 1 2  1Þ  tan1 M2  1  ðMÞ ¼ ¼ tan M ð γ1 γþ1 1 þ γ1 M2 M 2

(4.47f) Equation (4.47e) suggests that constant = 0 in Eqn. (4.47a) and that  is the total turning angle necessary for accelerating M ¼ 1 to a supersonic Mach number. Therefore,  is also called the Prandtl-Meyer angle. The numerical value of  ðMÞ is tabulated in Appendix C. For γ ¼ 1:4, this result is shown graphically in Fig. 4.1. The change in Mach number through an infinitesimal turning angle is found by combining Eqns. (4.46) and (4.47b):

dM γ1 2 dδ ¼ 1þ M pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (4.48a) M 2 M2  1 This is then used to obtain pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  M2  1 dM ¼ d dδ ¼ 1 þ γ1 M2 M 2

followed by integration, which yields δ¼R

(4.48b)

where R is a constant. Therefore  2 ¼  1 þ ðδ2  δ1 Þ and δ2 > δ1 > 0 suggest that  2 >  1 and M2 > M1 . Let Δ 12 ≡ ðδ2  δ1 Þ denote the turning angle. Then  2 ¼  1 þ Δ 12

(4.48c)

where Δ 12 > 0 results in Prandtl-Meyer expansion and Δ 12 < 0, oblique shock, consistent with the sign convention and results for oblique shock waves in Section 4.6. The use of Δ 12 in Eqn. (4.48c) indicates that the turning angle Δ 12 > 0 directly contributes to the Prandtl-Meyer angle for isentropic expansion. Δ 12 < 0 may be used to cause isentropic compression in the near-wall region of a gradually concaved surface such as the one depicted in Fig. 4.14c.

4.10 Expansion in Supersonic Flow by Turning The Prandtl-Meyer function shown in Eqn. (4.47f) has a zero value  ¼ 0 for M = 1. It cannot be defined for subsonic flows, and expansion by turning in subsonic flows is not possible. It can be seen from the table in Appendix C that as the turning angle away from the oncoming flow increases leads to dV > 0, which in turn leads

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4.10 Expansion in Supersonic Flow by Turning

93

to dM > 0 (i.e., M2 > M1 ). Such isentropic acceleration results in a decrease in pressure and thus flow expansion (p2 < p1 , according to Eqn. (4.16a)). Due to the isentropic nature of the process, both pt and Tt remain unchanged through the turning process: pt2 ¼ pt1 Tt2 ¼ Tt1 Due to the increase in supersonic value of M through expansion, μ decreases correspondingly and causes the Mach line to tilt, in addition to turning, in the downstream direction. Therefore, the waves diverge in the direction away from the turning corner, as shown in Fig. 4.15, decreasing possible gradients and further ensuring that the expansion is isentropic. When dealing with supersonic expansion, the flow can be viewed as if it were an M = 1 flow that has undergone the Prandtl-Meyer expansion through an angle  corresponding to the Mach number. Then the angle of the corner Δ contributes to further turning. Therefore, by treating  þ Δ as the total turning from the imaginary M = 1 flow, conditions after the expansion can easily be found. Figure 4.15b graphically demonstrates such an idea. The following example illustrates the calculation procedure. An air flow at M = 4, T ¼ 300 K, and p ¼ 101 kPa undergoes Prandtl-Meyer expansion by turning around a 10° convex corner. Find the flow speed and pressure both before and after the expansion.

EXAMPLE 4.11

Solution – Because  ¼ 0 for M = 1, the flow with M1 ¼ 4 can be imagined as having turned through an angle  1 ¼ 65:785° (the value can be found in Appendix C). Now  2 ¼  1 þ Δ ¼ 65:785° þ 10° ¼ 75:785° . Using the table in Appendix C again, M2 ≈ 4:88: (For comparison, Example 4.9 illustrates that an M1 ¼ 4 flow decelerates to M2 ¼ 3:233 by turning around a 10° concave corner. Isentropic relations are used to find pressures and temperatures.) p2 p2 pt2 pt1 1 1 ¼ ¼ ¼ 0:0022  1  0:0066 3 p1 pt2 pt1 p1 p2 ¼ 33:367 kPa (p2 < p1 – expansion results in a lower pressure) T2 T2 Tt2 Tt1 1 ¼ 0:729; T2 ¼ 218:61° ¼ ¼ 0:1735  1  0:2381 T1 Tt2 Tt1 T1 V 1 ¼ M1

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi γRT1 ¼ 4:0  1:4  287  300 ¼ 1; 388:8 m=s

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4 One-Dimensional Flow, Shock, and Expansion Waves

V2 ¼ M2

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi γRT2 ¼ 4:88  1:4  287  218:61 ¼ 1; 446:3 m=s

□

(V2 > V1 – flow acceleration due to supersonic expansion)

Similar to isentropic compression by turning, the lack of a length scale in the region near the corner requires that fluid and flow properties must be constant along the Mach line. Since these Mach lines diverge from the corner while undergoing supersonic expansion, the first and the last Mach lines define the extent of the expansion fan, wherein the flow and fluid properties undergo changes from uniform flow in region 1 to region 2. As depicted in Fig. 4.15a, the fan angle is 12 ¼ Δ 12 þ sin1

1 1  sin1 ¼ Δ 12 þ μ1  μ2 M1 M2

(4.49)

where the subscript “1.2” denotes from regions 1 to 2 and Δ 12 is the corresponding turning angle. EXAMPLE 4.12

Find the expansion fan angle for the flow configuration in

Example 4.11. Solution – 12 ¼ Δ 12 þ sin1

1 1  sin1 ¼ 10° þ 14:478°  11:825° ¼ 12:653° M1 M2

Comments – The fan angle is always greater than the turning angle because in addition to the turning, the Mach line is further inclined in the flow direction due to expansion □ and acceleration. Although the sharp turn and expansion depicted in Fig. 4.15a are not thermodynamically possible, their representation by a series of gradual turns is not unique. The development of the Prandtl-Meyer function does not assume a known degree of gradualness. Therefore, when dealing with supersonic expansion, one may treat the total turning angle as the only relevant parameter, as established by Eqn. (4.47f), while ignoring the effect of gradualness. The following example illustrates that the effect of one larger turning angle is the same as that achieved by two (which can be extended to many) small turnings with the same total turning. For the flow in Example 4.11, instead of expanding through a single 10° convex corner, allow the expansion to occur at first around a 6° convex corner, followed by a 4° convex corner. Find the final Mach number and the expansion fan angles emanating from these corners. EXAMPLE 4.13

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4.11 Simple and Non-Simple Regions

95

Solution – Again M1 ¼ 4 and  1 ¼ 65:785° .  2 ¼  1 þ Δ 12 ¼ 65:785° þ 6° ¼ 71:785° . Therefore, M2 ≈ 4:50.  3 ¼  2 þ Δ 23 ¼ 71:785° þ 4° ¼ 75:785. Therefore, M3 ≈ 4:88. 12 ¼ Δ 12 þ sin1

1 1  sin1 ¼ 6° þ 14:478°  12:840° ¼ 7:638° M1 M2

23 ¼ Δ 23 þ sin1

1 1  sin1 ¼ 4° þ 12:840°  11:825° ¼ 5:015° M2 M3

Comments – (1) Similar results for the Mach number cannot be obtained for turning around concave angles, as demonstrated by Example 4.9, where oblique shocks are not isentropic processes. Even with a very gradually curved concave corner, the Mach lines converge in the region away from the corner and the resultant shock wave causes a non-uniform flow downstream of the last Mach line (Section 8.7 describes how a non-uniform entropy field generates vorticity even though the incoming flow is uniform and free of vorticity). Prandtl-Meyer expansion is isentropic throughout the whole flow field. While the flow experiences changes in the fan region, the entropy remains the same everywhere in the flow, in both near and far fields. The total increase in Mach numbers is the sum of the increases due to expansion around two consecutive corners with smaller angles with the same total angle of turning. (2) The fan angles add up to be the same as that for a single 10° turn, although they do not have the same corner as their vertices. □ A curious question is: How large a turning angle of a convex corner is needed to accelerate the flow from M = 1 to ∞? Applying these two Mach numbers as the lower and the upper limits to Eqn. (4.47c) yields the turning angle needed,  max : sffiffiffiffiffiffiffiffiffiffiffi ! π γþ1 1 (4.50)  max ¼ 2 γ1 For γ ¼ 1:4;  max ¼ 134:454° . Since the Prandtl-Meyer expansion is isentropic, the isentropic relation, Eqn. (4.50), suggests that a supersonic flow into a vacuum (as in the case of a rocket exhaust in the outer space) would lead to a maximum flow turning for a rocket nozzle exit (use Fig. 4.9e as an example), Mach number Me ¼ 4 (i.e.,  e ¼ 65:785° ), Δ ¼ ð max   e Þ ¼ 68:669° ; that is, the exit flow turns through an angle 68:669° away from the nozzle axis.

4.11 Simple and Non-Simple Regions The Prandtl-Meyer function for isentropic compression and expansion waves discussed so far can be summarized as

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4 One-Dimensional Flow, Shock, and Expansion Waves

b

du

a f

M>1

d

y e

c

χ dl

Figure 4.16 Schematics illustrating characteristics and wave regions. (a) Simple wave regions are: a-d-f-g, c-e-d, downstream region of b-g-f, and downstream region of c-e-f. The non-simple wave region is where the two different families of characteristics intersect.

Non-simple wave region: d–e–f–g Simple wave regions: d–e–f–g, c–e–d downstream of b–g–f downstream of c–e–f

 ¼ 1  δ

(4.51)

where the + sign indicates the turning is away from the flow and thus an expansion, whereas the − sign indicates the turning is into the flow and thus a compression. Each of these waves (compression and expansion) are called simple waves, characterized by straight Mach lines with constant fluid and flow properties and by the simple relation, Eqn. (4.51), between δ and . Waves generated by adjacent walls may intersect (or interact), as illustrated in Fig. 4.16. These simple waves belong to one of the two families (+ or –) depending on whether they emanate from walls that lie to the left (for the “+” family) or right (for the “–” family) of the flow. In literature, these waves are also called the characteristic lines or simply characteristics. The –/+ characteristic is also called the η=ξ characteristic or the left-/right-running characteristic. The angle between the characteristic and the flow (i.e., streamline) is equal to the Mach angle, μ ¼ sin1 ð1=MÞ, where M is the local Mach number. For the right-running characteristics, emanating from the upper wall in Fig. 4.16, the turning angle (away from the flow and thus resulting in expansion) and the Prandtl-Meyer angle are related by Eqn. (4.48b):  þ δu ¼ Q ðright-running or þ characteristicsÞ

(4.52a)

Similarly along a left-running characteristic, a similar turning angle (away from the flow) results in compression  þ δl ¼ R ðleft-running or  characteristicsÞ

(4.52b)

where, in general δu ≠ δl and therefore Q≠R. In Eqn. (4.52) R and Q are different constants. For the supersonic flow over a thin airfoil shown in Fig 4.17, only leftrunning characteristics exist as the airfoil surface itself is a streamline. In the region where simple waves of the two opposite families interact, the flow is not simple and the relation between δ and  given by Eqn. (4.51) does not hold. As illustrated in Fig. 4.16, the characteristics in the non-simple region are curved and the local values

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4.11 Simple and Non-Simple Regions

97

y n (normal to streamline) n (–)characteristics (left-running or –) Flow direction μ μ

θ

(Streamline direction)

Figure 4.17 Flow region bounded by one wall surface generates only simple wave regions.

ξ characteristics (right-running or +)

Left-running η characteristics

Airfoil

χ

of  and δ (hence the Mach number and velocity) at intersections of the left- and right-running characteristics can be determined by solving the simultaneous equations (4.52a) and (4.52b). The flow field is thus determined by the method of characteristics (a similar method for one-dimensional unsteady flow will be discussed in Chapter 10). Use Eqn. (4.48) to calculate the Mach numbers after the turning around the corner described in Example 4.9.

EXAMPLE 4.14

Solution – Equation (4.48) is rewritten for approximation as

ΔM γ1 2 Δδ ¼ 1þ M pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M 2 M2  1 For M1 ¼ 4 and a single turn with Δδ ¼ 10° ¼ 0:1745, ΔM1 ¼ 0:757 and M2 ¼ M1 þ ΔM1 ¼ 3:243 (compared with the value in Example 4.9, M2 ¼ 3:229,which can be considered as the “exact” value for no approximation was made.) With two turns, with each of Δδ ¼ 5° ¼ 0:0873, ΔM1 ¼ 0:379, and M2 ¼ M1 þ ΔM1 ¼ 3:621. ΔM2 ¼ 0:329 and M3 ¼ M2 þ ΔM2 ¼ 3:292 (compared with M3 ≈ 3:36 in □ Example 4.9) Repeat finding Mach numbers after the expansion processes in Examples 4.11 and 4.13 using Prandtl-Meyer relations.

EXAMPLE 4.15

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4 One-Dimensional Flow, Shock, and Expansion Waves

Solution – Again, use

ΔM γ1 2 Δδ ¼þ 1þ M pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M 2 M2  1 Δδ ¼ þ10° ¼ þ0:1745, ΔM1 ¼ 0:757 and M2 ¼ M1 þ ΔM1 ¼ 4:757 (M2 ≈ 4:88 in Example 4.11) For expansion by turning Δδ ¼ 6° ¼ 0:105 followed by Δδ ¼ 4° ¼ 0:070, ΔM1 ¼ 0:455 and M2 ¼ M1 þ ΔM1 ¼ 4:455 (M2 ≈ 4:50 in Example 4.12) Following the turning with Δδ ¼ 4° ¼ 0:070, ΔM2 ¼ 0:357 and □ M3 ¼ M2 þ ΔM2 ¼ 4:812 (M2 ≈ 4:88 in Examples 4.11 and 4.13).

4.12 Reflected Shock and Expansion Waves So far the deflection of supersonic flow by a wall/surface is either accompanied by an oblique shock wave emanating from a concave corner or through an expansion wave emanating from a convex corner. In this section, reflection of waves from the wall and interaction of these waves are of concern. Figure 4.18a illustrates the phenomenon of reflection of an oblique shock wave, created by a frictionless wedge surface AB, which impinges on another surface CDE. The surface CDE is parallel to the free-stream flow (Region 1, with Mach number equal to M1 ). The wedge in inclined at an angle δ1 to the incident flow and an oblique shock is, for the present discussion, attached to the vertex of the wedge. The oblique shock wave, as shown by the line AD, reaches the inner surface of the supersonic inlet at point D. Downstream of the oblique shock wave AD (Region 2), the Mach number is M2 . It is assumed that M2 > 1 for this discussion (M2 < 1 is possible as Fig. 4.12 shows). The flow in Region 2 makes an angle δ2 ð¼ δ1 Þ with the surface CDE and, upon its impingement on CDE, the flow has to turn to be parallel to the surface. This means that the surface CDE acts as a wedge to the incident flow from Region 2 and an oblique shock wave (again, assumed to be attached) is formed emanating from point D. One expects that M1 > M2 > M3 , pt1 > pt2 > pt3 , and p1 < p2 < p3 as a consequence of non-isentropic compression by the original and the reflected oblique shocks. When the incident Mach number and the deflection angle permit formation of an attached reflected shock, such a reflection is called a regular reflection. The strength of reflection is defined by p3 =p1 ¼ ðp3 =p2 Þðp2 =p1 Þ, which is calculated once M1n and M2n become known. The reflected shock can be cancelled if it is not desirable, by turning the portion of the wall DE, shown in Fig. 4.13a, by an angle δ1 away from the incident flow in Region 2. By doing this as shown in Fig. 4.18b, wave cancellation or neutralization is achieved. If the combination of the incident oblique shock wave and the associated deflected flow angle (from Region 2 in Fig. 4.18a, for example) does not permit an attached oblique shock wave to form at the surface, the shock wave becomes normal

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4.12 Reflected Shock and Expansion Waves

99

(a) δ1 A δ1

B

θi

2

M1 > 1

3 M2

1 C

θr

M3

D

δ2

E

(b)

Figure 4.18 Reflection and cancellation of oblique shocks (parts a and b) and Mach reflection (part c, where the shock DG is the Mach shock wave).

δ1 δ1 2 M1 > 1

M2

1 D

C

δ1 E

(c)

F 2

M1 > 1 C

3

M2

1

Slip line M3

G

M4 D

4 E

to the wall at the wall and is curved away from the surface to be tangent to the incident shock wave, as illustrated in Fig. 4.18c. Such reflections are called Mach reflections and the shock wave DG is called a Mach shock wave. The Mach shock wave occurs because the combination of the incident Mach and the wedge angle does not allow an attached oblique shock wave at point D in Fig. 4.18c. In this case, the surface CDE acts as a blunt-nose body to the incident flow with M2 . The shock wave GF resembles that shown in Fig. 4.5, a bow shock. The shock strengths of DG and GF are therefore different, with M3 > 1 (downstream of the Mach shock wave DG) while M4 < 1 (behind the bow shock GF), and the discontinuities of flow properties exist across the interface of Regions 3 and 4. The interface is thus a slip line. In many occasions, such sharp discontinuities do not exist because of the varying curvature of the shock wave. Instead, a layer of finite thickness results across which some characteristic change in properties occurs. Because entropy is one of the properties

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4 One-Dimensional Flow, Shock, and Expansion Waves

that changes over the thickness, the layer is sometimes called the entropy layer. Because the Mach shock wave is not exactly normal to the wall in the region near point G, the flow in Regions 3 and 4 (in the entropy layer) in Fig. 4.18c is more complicated than that in regular reflection (e.g., Region 3 in Fig. 4.18c). The following example illustrates the simpler analysis that can be done for a regular reflection. Consider the flow depicted in Fig. 4.18a with for M1 ¼ 4:0 with T1 ¼ 216:7 K and p1 ¼ 19:1 kPa and δ1 ¼ 10° . Find the air speed, loss in stagnation pressure, shock strength of the reflection, and the reflected shock angle.

EXAMPLE 4.16

Solution – From Fig. 4.12, for M1 ¼ 4:0 with T1 ¼ 216:7 K and p1 ¼ 19:1 kPa and δ1 ¼ 10° , θi ≈ 23° . Therefore, M1n ¼ M1 sin23° ≈ 1:56. Therefore, using shock table or Eqn. (4.18) yields p2 p2  p1 ¼ 2:673 and β ¼ ¼ 1:673 p1 p1 pt2 pt2  pt1 ¼ 0:910 or ¼ 9% pt1 pt1 T2 ¼ 1:361 T1 M2n ¼ 0:681 rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi rffiffiffiffiffiffi T1 T1 1 ° ¼ 3:156 M2t ¼ M1t ¼ M1 cosθ1 ¼ 4 cos 23 1:361 T2 T2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ M2 ¼ 3:229 Therefore M2 ¼ M2t 2n So far the solution is the same as that shown in part (1) of Example 4.9. For the flow to be parallel to the lower wall, it must be turned through an angle of δ2 ¼ 10° . By again using Fig 4.12 for M2 ¼ 3:229 and δ2 ¼ 10° , θr ≈ 25:5° . Therefore, M2n ¼ M2 sin 25:5° ≈ 1:39. Using the shock table or Eqn. (4.18), one finds p3 p3  p2 ≈ 2:088 and β ¼ ≈ 1:088 p2 p2 pt3 pt3  pt2 ≈ 0:961 or ≈ 3:9% pt2 pt2

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4.13 Intersection of Shock Waves

101

T3 ≈ 1:248 T2 M3n ≈ 0:681 sffiffiffiffiffiffi sffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffi T2 T2 1 ° ≈ 2:609 ¼ M2 cosθr ¼ 3:229  cos 25:5 M3t ¼ M2t 1:248 T3 T3 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ M2 ≈ 2:696 Therefore M3 ¼ M3t 3n pt3 pt3 pt2 ¼  ≈ 0:961  0:910 ¼ 0:875 pt1 pt2 pt1 The stagnation pressure loss due to the incident and the reflected shocks is 12.5%, which translates into



pt3 pt1 pt3 Δpt ¼ pt3  pt1 ¼ pt1   1 ¼ p1   1 pt1 p1 pt1 ¼ 19:1 

1  ð0:875  1Þ  361:7 kPa 0:0066 Δpt ¼ 361:7 kPa

T3 T2   T1 ¼ 1:248  1:361  216:7 K ¼ 368:1 K T2 T1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi The speed in Region 3 is V3 ¼ M3 γRT3 ¼ 2:696  1:4  287  368:1 ¼ 1; 036:8 m=s The overall shock strength due to the reflection is p3 =p1 ¼ ðp3 =p2 Þðp2 =p1 Þ ¼ 2:088  2:673 ¼ 5:581 The reflected shock angle is θr  δ2 ≈ 15:5° . T3 ¼

Comments – The reflection is not specular, i.e., ðθr  δ2 Þ ≠ θi ; here θi > ðθr  δ2 Þ. This is mainly because M2 ≠ M1 and that the wave angle depends on the Mach number □ and the deflection angle.

4.13 Intersection of Shock Waves Figure 4.19a shows the intersection (or interaction) between two oblique shocks of equal strength (that is, caused by the same wedge angle). Because of the symmetry, the central streamline plays the identical role as the flat wall CDE in Fig. 4.18a.

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102

4 One-Dimensional Flow, Shock, and Expansion Waves

(a) δ M1

1

2

3

M3 δ

Figure 4.19 Interactions of two oblique shocks (a) with equal strength and (b) with unequal strengths. Note the slip line in the case of unequal strengths.

(b) δ1 δ1

θ1

1

2 M3

θ2

M2

M1

3

δ2 δ2'

M1

Slip line

3'

θ2' M ' 3 M2'

M1 1’

2' θ1'

δ1' δ1'

The analysis for flows in Regions 1 through 3 is the same as that for the flow in Fig. 4.18a. The intersection of two shock waves of different strengths is shown in Fig. 4.19b. 0 0 The deflection angles are δ1 ≠ δ1 and M2 ≠ M2 . The final results, downstream of the 0 0 reflected shocks, require that in Regions 3 and 3 (1) the pressures are equal, p3 ¼ p3 , and (2) the flow directions are the same. Other properties (Mach number, tempera0 ture, speed, and entropy) are not expected to be the same in Regions 3 and 3 , which are separated by the slip line (or the shear layer due to different flow speeds, or the entropy layer due to different values in shock strength). With the requirement of 0 p3 ¼ p3 , the final flow direction δ and p3 and other properties can then be determined, as demonstrated in the following example. Referring to Fig. 4.19b, consider air flow at M1 ¼ 4 and δ1 ¼ 5° 0 0 and δ1 ¼ 10 . Find θ1 , θ1 , θ2 , θ2 , δ (the final resultant turning angle) and all other 0 properties in Regions 3 and 3 . EXAMPLE 4.17 0

°

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4.13 Intersection of Shock Waves

103

Solution – For M1 ¼ 4 and δ1 ¼ 5° , θ1 ≈ 18° from Fig. 4.12. Therefore, M1n ¼ M1 sin18° ≈ 1:24. Therefore, using shock table or Eqn. (4.20) yields p2 p2  p1 ¼ 1:6272 and β ¼ ¼ 0:6272 p1 p1 pt2 pt2  pt1 ¼ 0:9884 ¼ 1:2% pt1 pt1 T2 ¼ 1:1531 T1 M2n ¼ 0:8183 rffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi rffiffiffiffiffiffi T1 T1 1 ° ¼ 3:543 M2t ¼ M1t ¼ M1 cosθ1 ¼ 4 cos18 1:1531 T2 T2 Therefore, M2 ¼ 3:64. 0 0 δ1 ¼ 10° leads to θ1 ¼ 23° , also from Fig. 4.12 (or Appendix E). Therefore, M1n ¼ M1 sin23° ≈ 1:56. Therefore, using the shock table or Eqn. (4.20) yields 0

0

p2 p  p1 ¼ 2:673 and β ¼ 2 ¼ 1:673 p1 p1 0

0

pt2 p  pt1 ¼ 0:910 t2 ¼ 9% pt1 pt1 0

T2 ¼ 1:361 T1 0

M2n ¼ 0:681 rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi rffiffiffiffiffiffi T1 T1 1 0 ° ¼ 3:156 ¼ M1 cosθ1 ¼ 4 cos23 M2t ¼ M1t 1:361 T2 T2 0

0

Therefore, M2 ¼ 3:229. 0 Because the flow in Regions 2 is turning through a larger angle upward than that in Regions 2 downward, it is reasonable to expect that the final flow is 0 turned upward from the incident flow. By guessing the value of δ, p3 , and p3 can be found. If their values are equal, then the guessed value of δ is the final flow 0 deflection angle and other properties in Regions 3 and 3 can be determined.

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104

4 One-Dimensional Flow, Shock, and Expansion Waves

f

b

h μ3 = sin–1(1/M3)

M2

α

M3 d

M1 > 1

a M4 e M5

c g

μ5 = sin–1(1/M5) i

Figure 4.20 Shock- and expansion wave systems formed by a supersonic flow approaching a finite wedge followed by an afterbody.

First guess δ ¼ 6° , then δ2 ¼ δ1 þ δ ¼ 5° þ 6° ¼ 11° for M2 ¼ 3:64, θ2 ≈ 23:0° . Therefore, M2n ¼ M2 sin23:0° ≈ 1:42 and pp32 ≈ 2:186 and pp31 ¼ pp32  pp21 ≈ 2:186  1:627 ≈ 3:557. 0 0 0 0 0 M0 2 ¼ 03:229, θ1 ≈ 20:0° . Therefore, M2n ¼ Then δ2 ¼ δ1 þ δ ¼ 0 10°  6° ¼ 4° for 0 0 p p p p M2 sin20° ≈ 1:10 and p32 ≈ 1:2450 and p31 ¼ p30  p21 ≈ 1:245  2:673 ≈ 3:352. 2 p3 ° p3 Therefore for δ ¼ 6 , p1 > p1 . Guess δ ¼ 5° , then δ2 ¼ δ1 þ δ ¼ 5° þ 5° ¼ 10° for M2 ¼ 3:64, θ2 ≈ 22:0° . Therefore, M2n ¼ M2 sin22:0° ≈ 1:36 and pp32 ≈ 1:991 and pp31 ¼ pp32  pp21 ≈ 1:991  1:627 ≈ 3:239. 0 0 0 0 For δ ¼ 5° , then δ2 ¼ δ1 þ δ ¼ 10°  50° ¼ 5° for M2 0¼ 3:229, θ2 ≈ 20:0° . 0 0 0 0 p p p p Therefore, M2n ¼ M2 sin21° ≈ 1:16 and p32 ≈ 1:403 and p31 ¼ p30  p21 ≈ 1:403  2 2:673 ≈ 3:750. 0 p Therefore for δ ¼ 5° , pp31 < p31 . The solution must be 5° < δ < 6° . One may assume the final solution is δ ¼ 5:5° . The best final value can be 0 found by further iterations. Once the value of δ is found, the answers for M3 , M3 and all other properties can then be found following similar procedures to those □ in Examples 4.9 and 4.16.

4.14 Interaction of Shock- and Expansion Waves In Section 4.6, the weakening of bow shock in the far field from the nose region of the wedge was discussed. The reason for weakening can be understood by considering a simpler attached oblique shock formed by a supersonic flow approaching a finite

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4.15 Applications of Shock- and Expansion Waves

105

wedge followed by an afterbody, sketched in Fig. 4.20. Assume that the finite wedge was symmetrical in shape and is place at an angle of attachment, α, to the free stream. As illustrated in Fig. 4.20, two attached oblique shock waves emanate from the vertex of the wedge, resulting in uniform flows in Regions 2 and 3. Due to different deflection angles, the shock strengths are different and M2 ≠M3 . If both M2 and M3 are > 1, the Prandtl-Meyer expansion will occur at corners c and d (i.e., the wedgeplate junctions). Because of the qualitative similarity between the flow over the upper- and lower surfaces, it suffices to discuss the effect of the upper surface of the afterbody. These expansion fans emanating from point d accelerate the flow velocity and decrease the pressure in the flow region bdf. As a consequence, the shock wave inclines gradually from point b to point f and approaches the Mach wave angle μ3 ¼ sin1 ð1=M3 Þ. The curved shock wave ceg can be explained in a similar manner (Thompson, 1972).

4.15 Applications of Shock- and Expansion Waves For a given free-stream Mach number M1 > 1, the larger the deflection (or wedge) angle is, the larger the shock wave angle (as Fig. 4.12 or Eqn. (4.27) shows), accompanied by the higher shock strength β, which is given by β¼

2 2 p2  p1 2γM1n γ  1 2γM1n 2γ ¼   ¼ p1 γþ1 γþ1 γþ1 γþ1

(4.53)

where subscripts 1 and 2 designate the free-stream and the post-shock conditions. The drag on the wedge due to (p2  p1 ) is called shock drag or supersonic wave drag, in addition to and different from skin/viscous drag and flow separation drag. Therefore, to minimize shock drag, it is desirable to reduce the shock wave angle. For a given free-stream Mach number, this can be achieved by reducing wedge angle and/or by avoiding a blunt nose. As shown in Example 4.9, turning through a series of small deflection angles results in a smaller β than a single large deflection with the same total turning angle. From the thermodynamic point of view, as little entropy increase as possible is desirable, which is suggested by Eqn. (4.36). Some examples illustrating supersonic wave drag are shown in Fig. 4.21. A diamond-shaped airfoil (symmetric with respect to its midsection as well as the chord line) at zero angle of attack, along with the surface pressure, is shown in Fig. 4.21a. Such an airfoil experiences drag force because the pressure in Region 2 is greater than that in Region 3, as the pressure distribution on the upper (and same on the lower) surface indicates that p2 > p1 due to the oblique shock and p3 < p2 due to the Prandtl-Meyer expansion around the convex corner at the midsection of the airfoil. In this case, the drag force per unit wing span on either the upper or the lower surface is D ¼ ð p 2  p3 Þ

c=2 cosδ



c sin δ ¼ ðp2  p3 Þ tan δ 2

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106

4 One-Dimensional Flow, Shock, and Expansion Waves

(a)

Expansion fan

Oblique shock 2

1

Oblique shock

3

δ δ

M1 > 1

4

p2 p4

p1 p3

(b)

Expansion fan Curved shock

Oblique shock

1 4 M1 > 1

p4

Figure 4.21 (a) and (b): Pressure distributions over symmetric airfoils in supersonic flows at zero angle of attack, explaining the non-zero supersonic wave drag. (c) Pressure distribution over a thin flat-plate at an angle of attack in a supersonic flow.

p1

(c) 1 2

M1 > 1

3

4

p2 p4

p1 p3

and the total drag is D ¼ ðp2  p3 Þc tan δ

(4.54a)

or alternatively, when expressed in terms of the airfoil thickness (t) D ¼ ðp2  p3 Þt

(4.54b)

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4.15 Applications of Shock- and Expansion Waves

107

where c is the chord length and t is the thickness at the midsection. This drag force exists in the absence of lift force, is independent of the viscous force, and is proportional to the midsection thickness, t. Equation (4.54) suggests that thin airfoils are desirable for minimizing drag in supersonic flows, the reasons being: (1) they result in small values of ðp2  p3 Þ because of smaller values of β, and (2) since D∝ tan δ ≈ δ, the thinner the airfoil the smaller δ and t, and thus D. A symmetric and curved airfoil at zero angle of attack is shown in Fig. 4.21b. The supersonic wave drag on this airfoil would require integration of the surface pressures on both upper and lower surfaces, whose geometry affects local slopes (and turning angles) on the surfaces and thus the surface pressure. For the shock wave to be attached, the half angle at the leading and trailing edges should be less than θmax (determined by Eqn. [4.27] or graphically by Fig. 4.12). An extremely thin airfoil may be approximated as a thin flat-plate, as shown in Fig. 4.20c. It has neither lift nor supersonic wave drag at zero angle of attack (α ¼ 0). For α ≠ 0, it is clear that the lift and drag forces per unit wing span are, respectively, L and D: L ¼ ðpL  pU Þc cos α

(4.55a)

D ¼ ðpL  pU Þc sin α

(4.55b)

where the subscripts L and U denote the lower and the upper surfaces, respectively. Here the angle of attack serves as the wedge angle (δ) for the flow along the lower surface and as the turning angle for Prandtl-Meyer expansion ( 1U ) on the upper surface. Unlike the symmetry seen in Figs. 4.21a and 4.21b, a slip line emanates from the trailing of the flat-plate with α ≠ 0. Slip lines (or more diffused entropy layers) are also expected for symmetric airfoils with α ≠ 0. For thin airfoils at a small angle of attack, isentropic compression and expansion theory can be used to relate lift and drag forces to the incoming Mach number and the angle of turning. For the flat-plate depicted in Fig. 4.21c, Eqn. (4.35) can be rewritten as 0 1 2 p  p1 B γM1 C ¼ @qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A⋅α p1 M2  1

(4.56)

1

Where the + sign is associated with the lower surface that causes compression while the − sign, with the upper wall that causes expansion. It is convenient to express pressure in terms of the pressure coefficient (Cp ): Cp ≡

p  p1 p  p1 2 p  p1 ¼ 1 γp1 2 ¼ 2 1 2 p1 γM 1 2 ρ1 V1 2 γRT1 V1

(4.57)

where the positive and negative signs are, respectively for the lower and the upper surfaces. It is similarly convenient to obtain the lift and the drag coefficients:

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108

4 One-Dimensional Flow, Shock, and Expansion Waves

CL ≡

 ðpL  pU Þc cos α  ¼ CpL  CpU cos α 1 2 2 ρ1 V1 c

(4.58a)

CD ≡

 ðpL  pU Þc sin α  ¼ CpL  CpU sin α 1 2c ρ V 2 1 1

(4.58b)

Substituting Eqn. (4.57) yields for small angle α ( cos α ≈ 1; sinα ≈ α): 4α CL ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M12  1

(4.59a)

4α2 CD ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M12  1

(4.59b)

Returning to the diamond-shaped airfoil at α ¼ 0 in Fig. 4.21a, the pressure coefficients on the surfaces in Regions 2 and 3 are, respectively, 2δ Cp2;3 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M12  1

(4.59c)

where the “+” sign is for Region 2 (compression) and the “−” sign is for Region 3 (expansion). Therefore, with t ¼ ctan δ

  1 2 ρ V ctanδ D ¼ 2ðp2  p3 Þt ¼ ðp2  p3 Þc tan δ ¼ Cp2  Cp3 2 1 1

4δ2 1 ρ1 V12 c ≈ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M12  1 2 Thus
t 2 D 4δ2 4  q q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ≈ 2 2 ρ1 V1 c M12  1 c M12  1

CD ¼ 1

(4.60)

4.16 Thin Airfoil Theory Combining Eqn. (4.59b) for a flat-plate and Eqn. (4.60) for a diamond-shaped symmetric airfoil, one obtains the drag coefficient for this airfoil with thickness-tochord ratio of t=c placed at an angle of attack of α to the free stream. There exists one more contribution to drag if the airfoil is not symmetric with respective to its chord, that is, if the camber is not zero, as shown in Fig. 4.22. The thin airfoil theory is developed in the following to take care of the effect of camber. Assuming that the upper and the lower surfaces of a thin airfoil is described by yU ¼ yU ð xÞ and yL ¼ yL ð xÞ, respectively. Then

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109

h(x)

αl (x)

α h(x)

yc(x)

yU(x) yL(x)

Camber line

=

α

+

+ yU(x)

x

x

x

x yL(x)

Figure 4.22 Decomposition of a thin, asymmetric airfoil geometry into a combination of three simpler geometric components: angle of attack, camber, and thickness of a symmetric airfoil. Because of the linear approximation used in the thin airfoil theory, the coefficients of pressure/lift/ drag of the airfoil is the sum of the respective coefficients of the three components.

110

4 One-Dimensional Flow, Shock, and Expansion Waves

2 dyU CpU ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 M1  1 dx

(4.61a)



2 dyL CpL ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  dx M12  1

(4.61b)

Because the camber line consists of the midpoints between yU and yL , the profile can be decomposed into the camber line yc ð xÞ and the symmetric thickness tð xÞ. Thus dyU d dh ½yc þ hð xÞ ¼ αl ð xÞ þ ¼ dx dx dx

(4.62a)

dyL d dh ½yc  hð xÞ ¼ αl ð xÞ  ¼ dx dx dx

(4.62b)

where αl ð xÞ ¼ α þ αc ð xÞ and αc ð xÞ is the local angle of attack of the camber line. For lift and drag coefficients, the lift and drag forces need to be calculated first. Then ðc   1 2 CpL  CpU dx L ¼ ρ1 V1 2 0 1 D ¼ ρ1 V12 2





 ðc dyL dyU CpL  þ CpU dx dx dx 0

Combining Eqns. (4.61) and (4.62) with these expressions yields  ð  ð

2 12 ρ1 V12 c 4 12 ρ1 V12 c dyc 2 L ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi αl ð xÞdx dx M12  1 0 M12  1 0 #   ð "  ð "



# 2 dh 2 2 12 ρ1 V12 c dyL 2 4 12 ρ1 V12 c
dyU 2 þ αl ð xÞ þ D ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx dx dx dx M12  1 0 M12  1 0 where the factor 4 accounts for the symmetry of the upper and lower surfaces. The average value over the chord length is defined as ð 1 c ð⋅Þ ¼ ð ⋅ Þdx c 0 Thus ðαc Þ ¼ 0 by its definition and ðαl Þ ¼ ðα þ αc Þ ¼ ðαÞ þ ðαc Þ ¼ α

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4.16 Thin Airfoil Theory

111

ðαl Þ2 ¼ ðα þ αc Þ2 ¼ α2 þ 2ααc þ α2c ¼ α2 þ α2c where ααc ¼ 0 because α is given and ααc ¼ ααc ¼ 0. It follows that ð L 4 1 c 4α ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi αl ð xÞdx ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi CL ¼ 1 2c c ρ V 2 1 1 M12  1 0 M12  1

D 4 1 CD ¼ 1 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2c ρ V 2 1 1 M2  1 c 1

ð c "
0

αl ð xÞ

2

(4.63a)

2 # dh þ dx dx

" 2 # 4 dh 2 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi α þ α2c þ dx 2 M1  1

(4.63b)

The following observations are made from Eqn. (4.63) for the thin airfoil theory. 1. The lift coefficient and the lift force are proportional to the angle of attack and do not depend on the camber and the thickness of the airfoil. 2. The drag coefficient and drag force have three contributing factors: First, the drag arises due to the lift, which is proportional to α2 . Second, camber contributes in proportion to α2c , which is always positive regardless of the sign of αc . Third, the  2 thickness also makes a positive contribution to the overall drag, with dh being dx  t 2 analogous to c for the symmetric diamond-shaped airfoil. These three drag forces are in addition to frictional or shear drag forces due to fluid viscosity. 3. Both large camber and large thickness are to be avoided for drag reduction. EXAMPLE 4.18 Consider the symmetric diamond airfoil depicted in Fig. 4.21a, with its upper front surface alignment parallel to the free stream and its half angle equal to δ ¼ 5° , as shown in the figure. Assume the free stream Mach number M1 ¼ 4 and P1 ¼ 20 kPa and a chord length of 0.6 m. Find CL and CD .

1

δ = 5°

2

4

M1 3

5

Solution – One way of finding CL and CD is to first find L and D as described by Eqn. (4.57). There are now, however, four surfaces to consider. Region 2 in the figure is not disturbed and p2 ¼ p1 .

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112

4 One-Dimensional Flow, Shock, and Expansion Waves

Region 3: M1 ¼ 4 and δ13 ¼ 10° leads to θ13 ≈ 23° and M1n ¼ M1 sin23° ≈ 1:56. Therefore, using the shock table or Eqn. (4.20) yields p3 ¼ 2:673 p1 pt3 ¼ 0:910 pt1 T3 ¼ 1:361 T1 M3n ¼ 0:681 sffiffiffiffiffiffi sffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffi T1 T1 1 ° ¼ 3:156 ¼ M1 cosθ13 ¼ 4 cos23 M3t ¼ M1t 1:361 T3 T3 Therefore M3 ¼ 3:229 and the corresponding Prandtl-Meyer angle is  3 ≈ 54° . Region 4:  2 ≈ 65:78°  4 ¼  2 þ Δ 24 ¼ 65:78° þ 10° ¼ 75:78° and M4 ≈ 4:88; p4 ¼ 0:0022 pt4



p4 p4 pt4 pt2 pt1 1 ¼    ¼ ð0:0022Þð1Þð1Þ ¼ 0:333 0:0066 p1 pt4 pt2 pt1 p1 Region 5: p5  5 ¼  3 þ Δ 35 ¼ 54° þ 10° ¼ 64° : Thus M5 ≈ 3:87 and ¼ 0:0079:

pt5 p5 p5 pt5 pt3 pt1 1 ¼    ¼ ð0:0079Þð1Þð0:910Þ ¼ 1:089 0:0066 p1 pt5 pt3 pt1 p1 Eqn. (4.57) states Cp ≡

p  p1 p  p1 2 p  p1 ¼ 1 γp1 2 ¼ 2 1 2 p1 γM ρ V V 1 2 1 1 2 γRT1 1

Using Eqn. (4.58) by incorporating the angle of the four surfaces, each of which has a length of c=2 cos δ, leads to

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4.16 Thin Airfoil Theory

113





2 p3  p1 1 p5  p1 1 cos2δ þ cos0 2 cosδ 2 cosδ p1 p1 γM12 



p2  p1 1 p4  p1 1  cos0  cos2δ 2 cosδ 2 cosδ p1 p1

CL ¼





2 p3  p1 1 p5  p1 1 CD ¼ sin2δ þ sin0 2 2 cosδ 2 cosδ p1 p1 γM1 



p2  p1 1 p4  p1 1  sin0  sin2δ 2 cosδ 2 cosδ p1 p1 1 1 1 Because 2δ ¼ 10° , ¼ 1:4; p2 ¼ p1 , p3pp ¼ 1:673, p5pp ¼ 0:089, p4pp ¼ 0:667 1 1 1 and M1 ¼ 4, CL ¼ 0:1072 and CD ¼ 0:018. Now if the thin-airfoil theory is used, α ¼ δ ¼ 5° ¼ 0:0873ðradiansÞ, dh 2 dx ¼ tan α ≈ α, αc ¼ 0 because of symmetry, and also αc ¼ 0 along the whole chord length.

4α 4  0:0873 CL ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:090 42  1 M12  1 " 2 # 4 dh 4 2 2 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ½0:08732 þ 0 þ 0:08732  ¼ 0:0158 CD ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi α þ αc þ 2 dx 4 1 M12  1

□

Consider a flat-plate that is at an angle of attack of α ¼ 10° to a free stream of M1 ¼ 4. (a) Find CL and CD considering the shock- and expansion waves. (b) Repeat for α ¼ 5° . (c) For both angle attacks also use the thin airfoil theory.

EXAMPLE 4.19

Solution – At the lower surface: (a) M1 ¼ 4 and δ1L ¼ 10° leads to θL ≈ 23° and M1n ¼ M1 sin 23° ≈ 1:56. Therefore, using the shock table or Eqn. (4.20) yields pL ¼ 2:673 p1 At the upper surface:  1 ≈ 65:78°  U ¼  1 þ Δ 1U ¼ 65:78° þ 10° ¼ 75:78° and MU ≈ 4:88 pptUU ¼ 0:0022

pU pU ptU pt1 1 ¼   ¼ ð0:0022Þð1Þ ¼ 0:333 0:0066 p1 ptU pt1 p1 Therefore,

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4 One-Dimensional Flow, Shock, and Expansion Waves

CL ¼

2 γM12

CD ¼

2 γM12





 pL  p1 pU  p1 cos α  cos α ¼ 0:206 p1 p1





 pL  p1 pU  p1 sin α  sin α ¼ 0:036 p1 p1

Simple geometric considerations also yield CD ¼ CL tan α. (b) M1 ¼ 4 and δ1L ¼ 5° leads to θL ≈ 18° and M1n ¼ M1 sin18° ≈ 1:24. Therefore, using the shock table or Eqn. (4.20) yields pL pL  p1 ¼ 1:6272 and β ¼ ¼ 0:6272 p1 p1 At the upper surface:  1 ≈ 65:78°  U ¼  1 þ Δ 1U ¼ 65:78° þ 5° ¼ 70:78° and MU ≈ 4:41

pU ¼ 0:0039 ptU



pU pU ptU pt1 1 ¼   ¼ ð0:0039Þð1Þ ¼ 0:591 0:0066 p1 ptU pt1 p1 Therefore, CL ¼

2 γM12

2 CD ¼ γM12





 pL  p1 p U  p1 cos α  cos α ¼ 0:092 p1 p1 



 pL  p1 p U  p1 sin α  sin α ¼ 0:008 p1 p1

The thin-airfoil theory yields δ ¼ 10° ¼ 0:1754 : 4α 4  0:1754 CL ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:180 42  1 M12  1 4α2 4  0:17452 CD ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:0314 42  1 M12  1 δ ¼ 5° ¼ 0:0873 :

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4.17 Prandtl’s Relation

115

4α 4  0:0873 CL ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:090 2 42  1 M1  1 4α2 4  0:08732 CD ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:0078 42  1 M12  1 Comments – Based on the results of Examples 4.18 and 4.19, the thin-airfoil appears to underpredict both CL and CD compared with the shock-expansion theory. As the angle of attack is reduced, the agreements between thin-airfoil predictions and □ shock-expansion results improve.

4.17 Prandtl’s Relation A useful Prandtl relation is now derived to directly relate velocities across normal or oblique shock waves without needing to know the Mach numbers. Equations (4.10) through (4.11) constitute the governing equations for a normal shock wave. With some algebraic manipulation for a normal shock, one obtains V1 V2 ¼ a2

(4.64)

where the superscript * denotes the sonic condition. For an oblique shock wave the Prandtl relation becomes V1n V2n ¼ a2 

γ1 2 V γþ1 t

(4.65)

It is noted that across an oblique shock the tangential velocity component does not change, that is, V1t ¼ V2t ¼ Vt . The derivation of Eqns. (4.64) and (4.65) is left as an exercise problem. When deriving Eqn. (6.45), attention should be paid to the fact that it is the normal component of the velocity that contributes the changes across the wave. The usefulness of Eqns. (4.64) and (4.65) can be seen in the following examples, and in a later chapter (Chapter 11) on traveling shock waves. For the same condition given in Example 4.3, use the Prandtl relation to calculate the air velocity downstream of the normal shock.

EXAMPLE 4.20

Solution – pffiffiffiffiffiffiffiffiffiffiffi For M1 ¼ 2:0 T1 ¼ 216 K, V1 ¼ γRT1 ¼ 590:2 m=s and



γ1 2 2  M1 T ¼ T t =ð γ þ 1Þ ¼ 2 T 1 þ ¼ 324 K: 2 γþ1

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116

4 One-Dimensional Flow, Shock, and Expansion Waves

a ¼ 361:4 m=s

V2 ¼

a2 ¼ 221:3 m=s V1

Comments – The result of V2 is the same as in Example 4.3, where the approach is intuitive.

□

Applying the Prandtl relation for the oblique shock problem of Example 4.6, find the downstream velocity and Mach number.

EXAMPLE 4.21

Solution – In Example 4.6, the given conditions are: air at 30 °C and traveling at Mach 4.0, experiencing an oblique shock at an angle θ ¼ 30° . Following results of Example 4.6, a1 ¼ 349:53 m=s M1n ¼ M1 sinθ ¼ 4 sin30° ¼ 2, V1n ¼ M1n a1 ¼ 699:1m=s, and T2 =T1 ¼ 1:6875 M1t ¼ M1 cosθ ¼ 4 cos30° ¼ 2:667, Vt ¼ V1t ¼ V2t ¼ M1t a1 ¼ 1210:8 m=s sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi



pffiffiffiffiffiffiffiffiffiffiffi γ1 2 γþ1   M1 = ¼ 653:9 m=s a ¼ γRT ¼ γRT1 1 þ 2 2 Therefore, Eqn. (4.65) gives ð699:1m=sÞV2n ¼ ð653:9m=sÞ2 

1:4  1 ð1210:8 m=sÞ2 ; 1:4 þ 1

V2n ¼ 262:1 m=s, and V2 ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ V 2 ¼ 1; 238:9 m=s V2n 2t

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M2 ¼ V2 =a2 ¼ V2 = γ  R  1:6875  ðT2 =T1 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ v2 = 1:4  288  1:6875  303 ¼ 2:73 These values of V2 and M2 are the same as for those obtained in Example 4.6. However, the approach used in Example 4.6 is intuitive, using shock relations and without the need for manipulating conservation laws for the Prandtl relation. However, the Prandtl relation is easy to use, as this example demonstrates. □

4.18 The Rankine-Hugoniot Equation The Prandtl relation directly relates gas velocities across shock waves without needing to know the Mach number. It is also desirable and advantageous to relate thermodynamic quantities across the shock bypassing the need for Mach number.

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4.18 The Rankine-Hugoniot Equation

117

The following derivation leads to the useful Rankine-Hugoniot relation that directly relates thermodynamic quantities across the shock. Recalling the conservation equations for mass and momentum in one dimension, one can rewrite them respectively as ρ1 V1 ¼ ρ2 V2

(4.66)

ρ1 V12 þ p1 ¼ ρ2 V22 þ p2

(4.67)

and

The energy equation is h1 þ

V12 V2 ¼ h2 þ 2 2 2

Noting that h ¼ cp T, Eqn. (4.68) becomes



p2 p1 2γ p2 p1 2 2 V1  V2 ¼ 2cp ðT2  T1 Þ ¼ 2cp   ¼ γ  1 ρ2 ρ1 ρ2 R ρ1 R

(4.69)

(4.70)

By combining Eqn. (4.66) and Eqn. (4.67), one finds

p2 p1 V1  V2 ¼  ρ2 V2 ρ1 V1 Multiplying this expression by ðV1 þ V2 Þ yields

1 1 2 2 V 1  V 2 ¼ ð p2  p 1 Þ þ ρ1 ρ2

(4.71)

Equating the right-hand sides of Eqns. (4.70) and (4.71) leads to the useful RankineHugoniot relation (also called the equation for a shock-adiabat curve for a perfect gas)
 γþ1 p2 v–2 p1 ¼ γ1  v–1
 (4.72) γþ1 v–2 γ1 v–1  1 or
 γþ1 p2 γ1 p1 þ 1 ρ2 ¼
 γþ1 ρ1 þ p2 γ1

(4.73)

p1

The derivation of Eqn. (4.72) is left as an exercise problem. Equation (4.73) is qualitatively sketched in Fig. 4.23. The upper branch of the Hugoniot curve (to the upper left of Point 1, the condition downstream of the shock wave) is represented by a solid line. For comparison, the isentropic p  –v relation is also presented, by a dashed line, due to the known (i.e., isentropic) thermodynamic

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118

4 One-Dimensional Flow, Shock, and Expansion Waves

p2 p1

M1> 1 M2< 1

Normal shock

M1= 1 (M2 = 1)

Isentropic

Figure 4.23 The Hugoniot curve (the shock adiabat) vs. the isentropic solution. The Hugoniot curve results from having to satisfy three conservation equations (mass, momentum, and energy), while the isentropic curve requires only conservation of energy and the isentropic condition.

1

1

γ –1 γ +1

υ2 ρ (= ρ1 ) υ1 2

path. It is noted that the key difference between the two curves in Fig. 4.23 is that the Hugoniot curve is the result of three conservation equations (mass, momentum, and energy), while the isentropic curve is obtained using only conservation of energy and the requirement that the process be isentropic. Because Eqns. (4.72) and (4.73) contain only thermodynamic quantities and they reduce to a function p2 ¼ p2 ð–v2 Þ or p2 ¼ p2 ðρ2 Þ, the function is called the shock adiabat (Shivamoggi, 1998; Thompson, 1972). Note that the Rankine-Hugoniot relation is valid for expansion as well as for compression. For very strong normal shocks (p2 =p1 ! ∞),

ρ2 γþ1 lim ¼ γ1 p2 =p1 !∞ ρ1 For a pressure ratio of 4.5 across a normal shock, find the temperature ratio using the Rankine-Hugoniot relation. Assume γ ¼ 1:4. Compare the result using the normal shock relationships or table.

EXAMPLE 4.22

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4.18 The Rankine-Hugoniot Equation

119

Solution – It is convenient to use Eqn. (4.72b) to find that vv––12 ¼ 2:667. By using the ideal gas law, T2 p2 – v2 1 ¼ 1:6875 ¼ ¼ 4:5  2:667 T1 p1 – v1 Alternatively,


2 γþ1 γþ1 p–2 p–2 p–2 γ1 p–1 þ p–1 T2 p2 – v2 p2 γ1 þ p–1 6  4:5 þ 4:52 ¼ 1:6875 ¼ ¼
 ¼
 ¼ γþ1 p–2 T1 p1 – v1 p1 γþ1 p–2 þ 1 6  4:5 þ 1 þ1 γ1 p–1

γ1 p–1

By using the following normal shock relation (or table) p2 2γM12 γ  1 ¼ 4:5  ¼ p1 γ þ 1 γ þ 1 one finds M1 ¼ 2:0 Then either by using the normal shock relation (or table in Appendix D), the temperature ratio across the shock is T2 ¼ 1:6875 T1

□

Use Eqn. (4.72a) to find a Hugoniot relationship for pressure ratio (p2 =p1 ) in terms of density ratio (ρ2 =ρ1 ). Use the result to find the percentage increase in pressure for a 2% increase in density.

EXAMPLE 4.23

Solution – v1 with ρ1 =ρ2 in Eqn. (4.72a) or rearranging Eqn. (4.72b) Either by replacing – v2 =– leads to
 γþ1 ρ2 γ1 ρ1  1 p2
 ¼ γþ1 p1  ρ2 γ1

ρ1

The percentage changes in density and pressure are, respectively, ðρ2  ρ1 Þ=ρ1 and ðp2  p1 Þ=p1 . Therefore,

 γþ1 ρ2 γþ1 ρ2 ρ1 γþ1 γ1 ρ1  1 γ1 ρ1  1 þ γ1 p 2  p1 p 2 ¼ 1¼
 1¼
 1 γþ1 γþ1 p1 p1  ρ2  ρ2 ρ1  1 γ1

ρ1

γ1

ρ1

For an x change in density,

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4 One-Dimensional Flow, Shock, and Expansion Waves


 γþ1 ð 1 þ x Þ γ1  1 p2  p1
 ¼ 2 p1 γ1  x In air ðγ ¼ 1:4Þ and x ¼ 0:002, ðp2  p1 Þ=p1 ¼ 0:00280 ¼ 2:8% .

□

4.19 Jump Conditions Because of the sharp gradients, the gas and flow properties are said to “jump” across the shock. It is useful to directly specify these jump conditions, by using ½Ψ ≡ Ψ 2  Ψ 1 to denote the difference between the downstream (subscript 2) and upstream (subscript 1) values of the parameter Ψ; the jump conditions are frequently used in literature. The one-dimensional continuity equation, for example, in terms of jump conditions is ½ρV ¼ 0, while for a one-dimensional nozzle flow where the flow crosssectional area varies (to be discussed in detail in the next chapter) the continuity is ½ρVA ¼ 0. It is left as an end-of-chapter exercise to rearrange Eqns. (4.22a), (4.22c), (4.22e) to arrive at the following jump conditions: 2 ¼ M2n

2 2 M1n þ γ1 2γ 2 γ1 M1n

1

2 2γM1n ½p ½p2 p1  γ1  ¼ ¼β¼ p1 p1 γþ1 γþ1

(4.74a)

(4.74b)



½V V2n V1n 2 1 M1n  ¼ ¼ a1 γþ1 M1n a1

(4.74c)



½– v – v2 –v1 2 1 1 2 ¼ ¼ v1 – γþ1 v1 – M1n

(4.74d)

The Rankine-Hugoniot relation can also be rearranged in terms of jump conditions as 2 ½p ½– v p1 ¼ –v1 2γ þ ðγ þ 1Þ ½p

(4.75)

p1

One useful relationship relating jump conditions of pressure and velocity is found by combining Eqns. (4.74b) through (4.74d) and Eqn. (4.75); it is left as an exercise at the end of the chapter:

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4.21 Hodograph Diagrams for Expansion Waves

2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3



½p γðγ þ 1Þ ½V 2 4 4 a1 2 5 ¼ 1þ 1þ p1 4 a1 γ þ 1 ½V

121

(4.76)

For normal shocks, results of Eqn. (4.74a) through (4.74d) are directly applicable by simply dropping the subscript n that denotes the normal component in case of oblique shocks.

4.20 Criterion for Strong Shocks So far a strong shock is associated with large values of M1 or M1n or β ≫ 1. Jump conditions should provide a more quantitative criterion. The criterion for strong and weak shocks in this section is adopted from Thompson (1972). Following the results of Problem 4.29, the non-dimensional pressure jump, denoted by ∏, is defined as ∏≡

½p ½p ¼ 2 ρ1 a1 γp1

(4.77)

which is equal to β=γ. Thus, the weak and strong shock approximation can be adopted, respectively, for ∏ ≪ 1 and ∏ ≫ 1.

4.21 Hodograph Diagrams for Shock- and Prandtl-Meyer Expansion Waves For one-dimensional oblique shock waves, Fig. 4.12 provides graphical relations among the free stream Mach number (M1 ), the turning angle (δ), and the shock wave angle (θ), which are visually easy to grasp. The graphical relations between upstream and downstream velocities can be achieved by plotting them on the hodograph plane, where the horizontal and vertical axes represent, respectively, the

V2y

V2, M2

V1, M1 q

δ

V 2t

2n

V

=V

1t

=V

1n

V

t

V2

q δ

q

V2y V2x

V2x V1x = V1

Figure 4.24 The velocity components across an oblique shock on the hodograph plane.

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122

4 One-Dimensional Flow, Shock, and Expansion Waves

downstream x- and y-velocity components. To this end, an oblique-shock velocity diagram can be constructed and is shown in Fig. 4.24. Note that in Fig. 4.23, Vt ¼ V1t ¼ V2t , and Vt ? V1n and V2n . The geometry indicates that V2y Vt ¼ V1n V1  V2x

(4.78a)

V2y Vt ¼ V1 V1n  V2n

(4.78b)

and

Combining with the energy equation, Eqn. (4.1), for a perfect gas and   ðh2  h1 Þ ¼ cp ðT2  T1 Þ ¼ cp ðγRT2  γRT1 Þ=γR ¼ a22  a21 =ðγ  1Þ, one arrives for normal shock waves V12 þ

a21 a2 γ þ 1 2 a ¼ V22 þ 2 ¼ γ1 γ1 γ1

(4.79a)

V2y a*

δmax Strong solution M2 =

θ

μ

* V 2/a

rarefraction /a*

M1 = V2

δ

V2x a* rarefraction Mach wave Oblique shock polar (weak solution)

M

2

=

V

2 /a

*

=1

Normal shock solution

Figure 4.25 Equation (4.80) is plotted as shock polar (solid lines) that represents weak solutions of the oblique shock equation. The dashed lines represent the rarefaction branches of the shock polar. The rarefaction branch is shown in further details in Fig. 4.26.

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4.21 Hodograph Diagrams for Expansion Waves

123

V2y /a*

γ +1 γ –1

M=

d M=

c’’

1

c *

e

ν max M1 f

=V

1

/a

rΔ

δo

ν ν

–Δ

b c’

ν

ν=0

a

V2x /a*

Figure 4.26 The rarefaction branch of the shock polar.

and when modified for oblique shocks 2 þ V1n

a21 a2 γ þ 1 2 2 a  Vt2 ¼ V2n þ 2 ¼ γ1 γ1 γ1

(4.79b)

After combining Eqns. (4.78a), (4.78b) and (4.79b) and some manipulations, the equation of the shock polar is obtained as V    2 V2x 2 V1 V2x 1 V2y a  a a a  1 ¼     2 V1 2 a  V1 V2x  1  γþ1 a

a

(4.80)

a

which relates V2y =a to V2x =a for the know value of V1 =a . Equation (4.80) is plotted and is qualitatively shown in Fig. 4.24. For given M2 and δ, there are two possible solutions: the weak solution and the strong solution. In Fig. 4.25, the shock polar represents weak solutions of the oblique shock equation, corresponding to the solid-line solutions. The Mach wave occurs when the supersonic flow encounters no obstacle (due to the absence of a twodimensional wedge in this case). The symmetry about the horizontal axis simply means that the deflection could be either upward or downward turning. Oblique shock results are represented by the curve to the left of the point of the Mach wave in Fig. 4.25. Recall that there are two different types of results downstream of oblique shock, represented by the strong and the weak solutions. The strong shock solution indicates subsonic downstream velocity (M2 ¼ V2 =a < 1), recalling that it is represented by dashed lines in Fig. 4.12. The weak solutions lead to mostly supersonic downstream conditions except for a narrow range of δ between the M2 ¼ 1 and the δmax lines in Fig. 4.12, corresponding to the similarly portion of the shock polar between the M2 ¼ 1 circle and the point denoted by δmax . The maximum turning angle (δmax ) for a given M1 results in an M2 value only slightly smaller than one,

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4 One-Dimensional Flow, Shock, and Expansion Waves

consistent with findings in Fig. 4.12. The normal shock solution is a special (and also the extreme) case on the shock polar, and is at the intersection of the shock polar and the V2x axis, with no y-velocity component as expected. The curves (represented by the dashed lines) to the right of the Mach wave are the rarefaction branches, due to Prandtl-Meyer expansion. While the shock polar correctly indicates that downstream velocity magnitude decreases as δ increases, the Prandtl-expansion branch indicates increasing velocity as δ increases. The details of the rarefaction branch is plotted in Fig. 4.26 using Eqn. (4.47c), given again below for quick reference: ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M2  1 dM γþ1 γ1 1 2  1Þ  tan1 M2  1 ¼ tan M ð  ðMÞ ¼ γ1 γþ1 1 þ γ1 M2 M 2

Figure 4.26 represents Mach number (M) as a function of the flow turning angle (δ and δ ¼ ) and demonstrates that the solutions for the rarefaction branch of the shock polar only exists for Mp>ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1, as the solution lies between the concentric circles ffi with radius equal to 1 and ðγ þ 1Þ=ðγ  1Þ, the latter being the maximum Mach number achievable by Prandtl-Meyer expansion, and for  ≤  max . The horizontal axis is for  ¼ 0, on which the hodographic solution is M ¼ 1 and the freestream velocity V1 =a > 1 shown in Fig. 4.25 is the point at the rarefaction branch with a corresponding Prandtl-Meyer angle . The dashed curve b  c00  e  f is the shock polar, same as that depicted in Fig. 4.25. Consider a flow with M1 > 1, denoted by point b on the rarefaction branch in Fig. 4.26. The flow slows down when  is decreased, moving along the rarefaction branch toward the horizontal axis ( ¼ 0) and toward M ¼ 1 (point a). By a decrease equal to Δ, the flow decelerates to point c’, with a smaller M. Such deceleration is isentropic, implied by the assumption for Eqn. (4.47c) and is indeed supersonic compression by isentropic turning discussed in Section 4.8. Similarly, increasing  from point b by Δ results in the flow represented by point c, corresponding to Prandtl-Meyer expansion to a higher Mach number. The symmetry of the rarefaction branches in Fig. 4.24 makes the conditions of points c0 and c identical. Turning the flow by a wedge angle of δ ¼ Δ from point b, however, leads to the oblique shock solution represented by point c00 on the shock branch (curve b  c00  e  f ) in Fig. 4.26. Points c00 , c, and d in Fig. 4.26 thus represent results of turning the a supersonic flow through an angle of Δ by a two-dimensional wedge, by isentropic compression, and Prandtl-Meyer expansion, with the resultant Mach numbers of Mc00 , Mc , and Md . By now it is understood that Mc00 < Mc < Md and this result is qualitatively shown in Fig. 4.26.

Problems Problem 4.1 Derive Eqn. (4.15). Problem 4.2 Derive Eqns. (4.16a), (4.16b), (4.16c), and (4.17b).

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Problems

125

Problem 4.3 Derive Eqns. (4.18a) and (4.18b). Problem 4.4 A converging-diverging nozzle made of unspecified materials and having unknown quality of surface finishing is attached to a reservoir of pressurized air. In the flow direction, three locations were selected for temperature and pressure measurements. The results are Location 1 – M1 ¼ 0:2, T1 ¼ 50° C, and p1 ¼ 3:00 MPa Location 2 – M2 ¼ 0:7, T2 ¼ 20° C, and p2 ¼ 2:13 MPa Location 3 – M3 ¼ 1:2, T3 ¼ 5° C, and p3 ¼ 1:20 MPa The ambient air has a temperature of 25° C. For air, cp ¼ 1:004 kJ=kg ⋅ K and R ¼ 0:287kJ=kg ⋅ K. Find Tt1 , Tt2 , pt1 , pt2 , the amounts of heat transfer to the air and entropy changes between locations 1 and 2 and between locations 1 and 3. Is the flow isentropic? Is the flow at all possible (or, in other words, could the measurement results be wrong)? Problem 4.5 Show that for a very weak shock wave 2 2  1 ¼ 1  M2n M1n

Consult the normal shock table to determine in what range of M1n this approximation is satisfactory. Problem 4.6 An air flow with a static temperature and pressure of 25 °C and 101.3 kPa, respectively, is moving at Mach 2.5 before passing a normal shock. Let subscripts 1 and 2 denote the states of air upstream and downstream of the shock. Find: (1) the air velocities upstream and downstream of the shock wave; (2) the pressure rise across the shock; (3) the pressure of air if it is slowed down isentropically to the same speed as with the shock; (4) the loss in stagnation pressure due to the shock. Problem 4.7 A supersonic flow at Mach 2.5 and with a static temperature and pressure of 25 °C and 101.3 kPa, respectively, is to be slowed down to achieve a static pressure of 1,450 kPa using a supersonic diffuser. Determine the diffuser geometry (e.g., the proportions of inlet, throat, and the exit cross-sectional areas). Problem 4.8 Use the Rankine-Hugoniot relation to show that ρ2 =ρ1 > 1 for a shock wave. Problem 4.9 Derive (4.22a)-(4.22f) for weak normal shock waves. Problem 4.10 Derive Eqns. (4.33a-c) from Eqn. (4.32) Problem 4.11 Designate the sonic condition, with the superscript *, as the reference condition for flows both down and upstream of a normal shock wave. By using the energy equation, Eqn. (4.6), show that

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126

4 One-Dimensional Flow, Shock, and Expansion Waves 2Tt (1) T1 ¼ T2 ¼ γþ1

(2)

V12 2

a2

1 þ γ1 ¼

V22 2

a2

γ1 2 2 2 2 2 2 þ γ1 ¼ constant ¼ 12 γþ1 γ1 a or a ¼ γþ1 V1 þ γþ1 a1 a2

a2

2 2 2 2 2 1 (3) For an oblique shock: V1n þ γ1 ¼ V2n þ γ1 ¼ γþ1 γ1 a  Vt

Problem 4.12 (a) Using the energy equation, Eqn. (4.6), for a perfect gas flow across a normal shock, show that V1 V2 ¼

2 2 a γþ1 t

or V1 V2 ¼ a2 where subscripts 1 and 2 designate the upstream and the downstream regions, respectively. The above two relations are the Prandtl relation. This result makes readily clear that for M1 ¼ v1 =a > 1 (corresponding to M1 > 1, because T  > T1 and a > a1 ), M2 < 1 (corresponding to M2 < 1), as explained in Section 4.4. Then use the limiting case where M1 ¼ M2 ¼ 1 to show that sffiffiffiffiffiffiffiffiffiffiffi 2 at a ¼ γþ1 

which can also be alternatively shown in Problem 5.13. (b) Then show that for an oblique shock wave V1n V2n ¼ a2 

γ1 2 V γþ1 t

Problem 4.13 For the flow depicted in Fig. 4.17 find the reflected shock angle for M1 ¼ 2:0 and 1.6, with δ1 ¼ 10° in both cases. Compare with the results of Example 4.16. Problem 4.14 Show that for an isentropic flow of perfect gas dp=p γ ¼ : dT=T γ  1 That is, the rate of change in pressure is larger than that in temperature for a given Mach number as the flow accelerates or decelerates. Problem 4.15 Use energy conservation to show that the sonic (or critical) velocity, Vc ð¼ a Þ, and the maximum (or limiting) velocity, Vl , to obtain for an ideal gas in tank are, respectively,

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Problems

127

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffi 2γ pt 2 at ¼ V c ¼ a ¼ γ þ 1 ρt γþ1 and sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffi 2γ pt 2 at Vl ¼ ¼ γ  1 ρt γ1 Also show that at any point in an isentropic flow, the velocity V is vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi " # u γ1 u 2γ pt p γ t V¼ 1 γ  1 ρt pt Problem 4.16 Show that across a stationary normal shock wave (with subscripts 1 and 2 denoting upstream and downstream states of the shock, respectively) 1 (1) V1n V2n ¼ pρ2 p ρ 2

(2) J ¼ 2

(3)

p2 p1 ρ1 a21

1

1  pv22 p v1


 V2n V1n 2 v2 v1 1 ¼ M1 V2aV or ¼ M ¼ M 1n 1n a v 1 1 1

(4) ðV2n  V1n Þ2 ¼ ðp2  p1 Þðv2  v1 Þ where J ¼ ρ1 V1 ¼ ρ2 V2 is the mass flux across the shock wave and M1n , V1n , and V2n are used for an oblique shock; drop the subscript n for a normal shock. Problem 4.17 Use the isentropic relationship  γ pt γ  1 2 γ1 M ¼ 1þ 2 p to show that (1) At low Mach numbers the local static pressure can be expressed as p ¼ pt 

  ρV 2 1 2γ 4 M þ… 1 þ M2 þ 4 24 2

(2) The velocity determined by Pitot measurement by neglecting the Mach number, or compressibility, effect, has an error of

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128

4 One-Dimensional Flow, Shock, and Expansion Waves

 γ1 γ 1 pt 1 : 4ðγ  1Þ p∞ where p∞ is the free-stream static pressure. 2  1 ¼  ≪ 1), find ðpt2  pt1 Þ=pt1 in terms Problem 4.18 For a weak shock wave (M1n  2  of M1n  1 and M2n in terms of M1n . Assume a constant specific heat ratio, γ.

Problem 4.19 Can the entropy change across a shock wave be written as the following expression? s2  s1 cp T2 p2 cp Tt2 pt2 ¼ ln  ln ¼ ln  ln R R T1 p1 R Tt1 pt1 where subscripts 1 and 2 denote, respectively, upstream and downstream of the shock wave. Hint: Entropy is a thermodynamic property and there exists a stagnation entropy. Problem 4.20 Combine continuity, momentum, and energy considerations to show that for a normal shock in a general fluid that is not an ideal gas: 1 ρ1 þ ρ2 ðp2  p1 Þ 2 ρ1 ρ2

h2  h1 ¼

where subscripts 1 and 2 denote, respectively, upstream and downstream of the shock wave. Problem 4.21 Find p2 =pt2 downstream of a normal shock wave in terms of M1 , the upstream Mach number. Also show that h p1 ¼ pt2

1 iγ1  γ1 γþ1 γ h iγ1 γþ1 2 M 1 2

2γ 2 γþ1 M1

which is the Rayleigh supersonic Pitot formula that allows determining M1 by measuring pt2 that is the reading by the Pitot probe in supersonic flow, unlike pt1 read by the subsonic Pitot probe. Problem 4.22 A normal shock occurs standing immediately in front of a pitot-static probe placed facing and parallel to a supersonic air flow. Upstream of the shock the static pressure and temperature are measured to be p1 ¼ 11:0 kPa and T1 ¼ 200 K. Downstream of the shock the measured values are p2 ¼ 54:8 kPa and pt2 ¼ 68:1 kPa. Find V1 . Demonstrate that pt2 ¼ 68:1 kPa is a reasonable reading. Problem 4.23 Derive the Rankine-Hugoniot relationship between pressure and specific volume, Eqn. (4.72a), and then show that

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Problems

129

(1) it can be rewritten as Eqn. (4.72b), and (2) for a very weak shock wave, it is reduced to the isentropic relationship dp dρ p2 ¼γ or ¼ p ρ p1

γ ρ2 ρ1

(3) For a weak shock in air that causes a 0.2% increase in density, what is the corresponding increase in pressure? Compare this result with that in Example 4.23. Problem 4.24 A supersonic air flow (M ¼ 3:0) is going through a centered expansion fan through an 8° turning and then returns to the original flow direction through an oblique shock, as here shown. (1) Find the Mach number and pressure in each flow region shown in the figure using the shock/expansion theory, and (2) Use the Prandtl-Meyer isentropic turning (for both expansion and compression) to find the Mach number and pressure in region 3; compare these results with those of (1).

p1 M1 = 3.0

1 2 3

8º

Problem 4.25 The same supersonic flow is going through the expansion and oblique shock in the reverse order as those in Problem 4.24, as here shown. Find the Mach number and pressure in regions 2 and 3, and compare with the results of Problem 4.24. p1 M1 = 3.0

1

3

2 8º

Problem 4.26 A thin flat-plate airfoil with a trailing-edge flap as shown is placed in a uniform supersonic air stream. Both angles of attack and the flap deflection are small (i.e., α ≪ 1 and φ ≪ 1). Find the drag coefficient using the thin-airfoil theory. α

M1 c

Problem 4.28 A thin airfoil with flat surfaces is placed in a uniform supersonic flow with a Mach number M, as shown in the figure. Assuming 2t=c ≪ 1, find the drag coefficient. α M>1

α C/2

t

α C/2

Problem 4.29 A thin flat-plate is placed in an air stream with M ¼ 1:6, at an angle of attack of 8° . Find the lift and drag coefficients. Problem 4.30 A supersonic air flow (p ¼ 100 kPa and M ¼ 2:0) is deflected by a twodimensional wedge through an angle of 10° . Find the pressure and entropy change due to the deflection using (1) oblique shock theory and (2) isentropic turning. Compare the results. Problem 4.31 Derive Eqn. (4.74). Problem 4.32 Use the Rankine-Hugoniot equation to show the following jump conditions without invoking the Mach number: 2 ½p ½ρ p1 ¼ ρ1 2γ þ ðγ  1Þ ½p p1

2 ½p ½– v p1 ¼ –v1 2γ þ ðγ þ 1Þ ½p p1

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Problems

131

Problem 4.33 Combine Eqns. (4.74b) through (4.74d) and Eqn. (4.75) to show that 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3



½p γðγ þ 1Þ ½V 2 4 4 a1 2 5 ¼ 1þ 1þ p1 4 a1 γ þ 1 ½V Problem 4.34 Use the jump conditions, Eqn. (4.74) to show the following: (1) V1 V2 ¼ ½p ½ρ (2) J 2 ¼  ½p ½– v , where J ¼ ρ1 V1 ¼ ρ2 V2 is the mass flux across the shock wave (3)

½p ρ1 a21

– 2 ½v ¼ M1n ½V a1 ¼ M1n v–1

(4) ½V2 ¼ ½p½– v Problem 4.35 A shock wave system resulting from supersonic flow and wave interaction is described in the figure below. Find out the values of the following parameters – M3 , V3 , p, T3 , and pt3 . Problem 4.36 On a hodograph plane qualitative locate and explain the results of a supersonic flow undergo turning by a two-dimensional wedge through an angle of 2δ in two scenario: (i) by one single turning of 2δ and (ii) by two successive wedges each having a turning angle of δ. Also locate the result with a normal shock. Assume attached oblique shocks. For illustration, use the freestream Mach number M1 ¼ 4 and δ ¼ 5° on the hodograph plane (the numerical results are readily available from Examples 4.9, 4.10, and 4.11).

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5

Steady One-Dimensional Flows in Channels

Chapter 4 is mainly concerned with shock and expansion waves in external flows; it has also discussed the concept of supersonic inlet flows that can be slowed down either by a normal shock, accompanied by a loss in stagnation pressure, or a series of weaker, oblique shocks with a smaller loss in stagnation pressure. Flow in nozzles with a varying cross-sectional area was also mentioned in Section 4.19. In fact a supersonic flow can be slowed down without any shock waves by using a properly designed diffuser, which has a decreasing cross-sectional area in the flow direction. A diffuser is a nozzle in reverse, with a nozzle having an increasing cross-sectional area in the flow direction. One might therefore expect a nozzle to be used to accelerate supersonic flows, in applications such as rockets where exit flows should be as supersonic as allowed by the rocket chamber design. Both diffuser and nozzle flows are channel flows. For best performance, it is a general rule to avoid shock waves in the channel. This chapter addresses flows in channels, including supersonic wind tunnel operation and design. Conditions under which shock wave might occur in channels are of practical interest to thrust generation in propulsion applications.

5.1 Isentropic Flow of an Perfect Gas in Channels of Variable CrossSectional Area Consider a steady one-dimensional flow in a channel of variable cross-sectional area, schematically shown in Fig. 5.1. To focus on the effect of area change, assume (1) the flow is one-dimensional, (2) the flow is steady, (3) effects of viscosity and gravity are negligible, (4) there is no heat transfer through the channel wall (i.e., the wall is adiabatic). As a result of these assumptions, the flow is everywhere isentropic except where shock waves occur. The conservation of mass requires that ρVA ¼ m_ ¼ constant

(5.1)

Differentiating both sides of this equation and dividing through by ρVA yields

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5.1 Isentropic Flow in Variable Cross-Sectional Area M V p T

M + dM V + dV p + dp T + dT ρ + dρ

ρ

(p+dp) (A+dA)

pA

133

Figure 5.1 Schematic of a steady onedimensional flow in a channel of variable cross-sectional area.

dA (p+ dp )dA 2

dρ dV dA þ þ ¼0 ρ V A

(5.2)

The momentum conservation is governed by Euler’s equation, Eqn. (4.11): dp þ ρVdV ¼ 0

(5.3)

The energy conservation is given by Eqn. (4.6) 1 ht ¼ cp Tt ¼ h þ V 2 ¼ constant 2 In differential form, it becomes

2

1 2 V d h þ V ¼ dh þ d ¼0 2 2

(5.4)

Note that Reynolds transport theorem applied to the control volume shown in Fig. 5.1 will also lead to the same differential forms of the governing equations (5.2) through (5.4). This is left as a problem at the end of this chapter. Isentropic flows are reversible, and therefore Tds ¼ dh 

dp ρ

(2.36) (5.5)

and because ds ¼ 0 dh ¼

dp ρ

Combining this expression with Eqn. (5.4) leads to an equation identical with Eqn. (5.3); that is, for an isentropic flow, the momentum and energy considerations results in the same equation. It is rewritten as dp ¼ ρVdV or

dV dp ¼ 2 V ρV

(5.6)

Substituting this expression into the mass conservation relation, Eqn. (5.2), yields

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134

5 Steady One-Dimensional Flows in Channels

  dρ dA dp þ ρV 2   ¼0 ρ A

(5.7)

Because a2 ¼ ð∂p=∂ρÞs , dρ ¼ dp=a2 , which upon substituting into Eqn. (5.7) yields   dp dA dp þ ρV 2  2  ¼0 ρa A   dA dp 1  M2 ¼ ρV 2 A

(5.8a)

For a one-dimensional flow, A is a function only of the flow direction x, and Eqn. (5.8a) can be written as dp ρV 2 dA ¼ dx Að1  M2 Þ dx

(5.8b)

Again by using dp ¼ ρVdV for dp, one obtains 

M2  1

 dV dA ¼ V A

(5.9)

  For supersonic flows, M2  1 > 0 and, therefore, an increase in the crosssectional area (i.e., dA > 0) leads to flow acceleration (dV > 0). According to Eqn. (5.6), an increase in the cross-sectional area also causes dp < 0, a decrease in pressure. As the flow accelerates, (d½V 2 =2 > 0), and conservation of energy requires that dT ðor dhÞ < 0 according to Eqn. (5.4), which causes a to also decrease. These results for supersonic flows are in contrast to those in subsonic channel flows. In   subsonic flows, M2  1 < 0, dA > 0 causes dV < 0 (i.e., deceleration), dp > 0 and dT > 0; for dA < 0, dV > 0 (i.e., acceleration), dp < 0. The pressure variation with velocity discussed is thus consistent with the Bernoulli principle, whether M < 1 or M > 1. These results are summarized in Fig. 5.2. The results in Fig. 5.2 indicate that a subsonic flow cannot be accelerated to be supersonic in a converging nozzle and that a supersonic cannot be decelerated to be

Subsonic Flows, M1

dA < 0

dA > 0

dA < 0

dA > 0

dV > 0

dV < 0

dV < 0

dV > 0

dM > 0

dM < 0

dM < 0

dM > 0

dp < 0

dp > 0

dp > 0

dp < 0

Figure 5.2 Variations in flow velocity (V), Mach number (M), and pressure (p) in a steady one-dimensional flow in a channel of variable cross-sectional area.

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5.2 Reference Parameters in Nozzle and Diffuser Flows (a)

M1

M>1

135

dA = 0

M=1

M 0 for dA < 0) occurs in the converging section and supersonic acceleration (dV and dM > 0 for dA > 0) in the diverging section. Conversely, as shown in Fig. 5.3b, subsonic deceleration occurs in the diverging section and supersonic deceleration in the converging section. In other words, M is not a single-value function of A=A . However, both p and T are singlevalued functions of M, and they both decrease with increasing M. A converging-diverging nozzle is connected to a reservoir containing air at 300 kPa (pr ) and 300 K (Tr ).

EXAMPLE 5.1

(a) If air exits the nozzle at a pressure equal to that in the ambient (pb , also called the “back” pressure) 101.3 kPa, what is the ratio of cross-sectional areas of exit to the throat? Under this condition, what is the exit velocity? (b) What is the ambient pressure to cause a normal shock wave at the exit plane of such a nozzle? What is the exit velocity? (c) What is the value of pb for subsonic exit condition with Mth ¼ 1 and without shock, given the nozzle determined in (a)?

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5.2 Reference Parameters in Nozzle and Diffuser Flows

137

Solutions – (a) Assume the flow within the nozzle is isentropic. pb pe pe 101:3 Therefore, Me ≈ 1:34 and AA ≈ 1:084 and pt ¼ pt ¼ pr ¼ 300 ¼ 0:338. T T Tt ¼ Tr ¼ 0:7358 and T = 220.7 K. Because Me > 1, it follows that Mth ¼ 1, and Ath ¼ A and therefore A Ath ≈ 1:084. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi Ve ¼ Me γRTe ¼ 1:34  1:4  287  220:7 ¼ 399:1 m=s (b) Normal shock wave at the exit plane Let subscripts 1 and 2 designate, respectively, conditions upstream and downstream of the normal shock wave. Then M1 ¼ 1:34 and therefore M1 ≈ 0:77, pt2 p2 T2 pt1 ≈ 0:972. Downstream of the normal shock wave pt2 ≈ 0:675 and Tt2 ≈ 0:894. Thus, pb ¼ p2 ¼ ppt22  ppt2t1  pt1 ¼ 0:675  0:972  300 kPa ¼ 196:8 kPa. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi Ve ¼ V2 ¼ M2 γRT2 ¼ 0:77  1:4  287  0:894  300 ¼ 252:8 m=s (c) For AAth ≈ 1:084, Subsonic value of Me is Me ¼ M2 ¼ 0:715, for which pe Te Tt ¼ 0:907 and pt ¼ 0:710. pe ¼ pe ¼ 0:710  300 kPa ¼ 213:0 kPa Ve ¼ V2 ¼ M2

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi γRT2 ¼ 0:715  1:4  287  0:907  300 ¼ 236:4 m=s

Comments – Because ppet monotonically decreases with increasing M (see Eqn. (4.9) Appendix C), unlike AA , Me ≈ 1:34 is the only solution, for which AA is determined. For A A ≈ 1:084 a subsonic value Me ≈ 0:715 is a possible solution. But in this case pe pt ¼ 0:708, i.e., pe ¼ 212:4 kPa. For this to occur, however, pb is 213.0 kPa because a subsonic exit flow can adjust to pb . Deviation of pb from 213.0 kPa would result in different exit Mach number; such a scenario will be discussed □ further in section 5.3. Example 5.1 provides a scenario where a flow resumes subsonic conditions (and deceleration) in the diverging section after having accelerated from subsonic conditions in the converging section to reach sonic conditions at the throat. Whether the existing flow is supersonic or subsonic, because Me ¼ 1, the throat condition serves as a reference and specifically Ath ¼ A . On the other hand, A as a reference parameter might not physically exist for a given isentropic channel flow; the following example illustrates this situation. A converging-diverging nozzle with the exit to throat area ratio ¼ 1:084 is connected to a reservoir containing air at 300 kPa and 300 K. If the air exits the nozzle to another reservoir at a pressure of 270 kPa, what is the Mach number entering the second reservoir?

EXAMPLE 5.2 Ae Ath

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138

5 Steady One-Dimensional Flows in Channels

Solution – Assume an isentropic flow throughout the flow. Therefore pt ¼ pr and ppet ¼ pper Ae Ae ¼ 270 300 ¼ 0:9. Therefore Me ≈ 0:46 and A ≈ 1:425. Comparing with Ath ¼ 1:084, Ath  1:425 A < Ath and Mth ≠1. Because A ¼ 1:084 ≈ 1:315, Mth ≈ 0:51. Comments – (1) For AAth ≈ 1:315 the supersonic Mach number is Mth ≈ 1:68. However, the supersonic cannot exist at the throat, because Mth can at most be equal to unity. Thus, in this problem A does not physically exist and Mth < 1. (2) For exactly the same nozzle as that in Example 5.1, the deviation from pb ¼ 213:0 kPa (now 270 kPa) results in vastly different (sonic vs. subsonic) flow □ conditions at the throat. In dealing with supersonic jet noise, a fully expanded Mach number, MFE , is an important reference parameter. MFE is defined as the Mach number achieved by expanding the supersonic jet to match the surrounding pressure; that is, the value of MFE might not exist within the nozzle flow field. For an air reservoir pressure of 20 atmospheres, find the value of MFE for expansion into a surrounding having a pressure of 0.8 atmospheres.

EXAMPLE 5.3

Solution – pt 20 ¼ 25 ¼ ¼ p∞ 0:8



γ  1 2 γ=ðγ1Þ 1þ MFE 2

By solving the above expression, MFE ¼ 2:746. Comments – For flight vehicle, the nozzle exit plane pressure is usually higher than that of the surrounding (under-expansion instead of over-expansion to avoid shock waves within the nozzle or at the nozzle exit, both phenomena to be discussed in the next section). Thus, the reference value of MFE is not achieved within the nozzle. □

5.3 Standing Normal Shock Wave in Converging-Diverging Nozzles For applications such as rocket propulsion and gas delivery devices, the conditions at the exit of the channel are important. Two questions arise as to what exit conditions are if the ambient pressure (also called the back pressure pb ) falls between those for Me > 1 and Me < 1 (as illustrated in Example 5.1) but with Mth ¼ 1: (1) Is the back pressure equal to or lower than the pressure of the existing flow, and (2) Is the back pressure higher than the exit plane pressure? The steady-state momentum equation dp ¼ ρVdV indicates that as the flow accelerates (dV > 0) or decelerates (dV < 0), the static pressure decreases (dp < 0Þ or increases (dp > 0) correspondingly. This is true whether the flow is supersonic or subsonic, as no such distinction is needed for applying the principle of conservation

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5.3 Standing Normal Shock Wave in Nozzles

139

Throat (*) Pr Reservoir

P M 1

Mth < 1 Me < 1 M < 1 throughout the nozzle

2 3 4

Normal shock within nozzle

A B

M=1 (p=p*)

Me < 1

C

( )

*

D 5

Oblique shock outside nozzle

5

M=1 (p=p*)

Over-expanded nozzle

Choked nozzle (Mth =1)

4 Me > 1 3 2

6 (expansion wave outside nozzle)

Under-expanded nozzle; expansion outside nozzle

1

Figure 5.4 Scenarios of flow in a converging-diverging nozzle attached to a reservoir having a pressure pr . Note that the back pressure (pb ) or the ratio pb =pr plays an important role. For pb < p4 , the nozzle is choked with Mth ¼ M ¼ 1 at the throat. Perfect expansion occurs when pb ¼ p5 . For pC < pb < p4 , normal shocks exist in the diverging section, while for p5 < pb < pC , oblique shocks extends downstream of the nozzle exit and are attached to the nozzle lip.

of momentum. The momentum conservation principle can be used to gain knowledge of the pressure variation throughout a converging-diverging nozzle, including the exit plane, to answer the aforementioned two questions. Consider in Fig. 5.4 a converging-diverging nozzle attached to a reservoir having a pressure pr , with the profiles of pressure ratio p=pr and Mach number plotted along the flow direction. For pb =pr ¼ 1, dp ¼ ρVdV ¼ 0 everywhere within the nozzle; there is thus no flow and no pressure variation in the nozzle. This condition is

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140

5 Steady One-Dimensional Flows in Channels

described by the horizontal line (curve 1). As pb =pr is decreased below unity, a subsonic flow is induced at the entrance to the nozzle. For values of pb corresponding to curves 2 and 3, subsonic flows exist throughout the nozzle, including the throat. Under conditions represented by curves 2 and 3, the flow accelerates (Mach number increases and the pressure decreases) in the converging section and decelerates (Mach number decreases and the pressure increases) in the diverging section, with minimum pressure and maximum Mach number at the throat. Because pb is smaller for curve 3, the flow speed and Mach number are higher and the pressure is lower at same locations (i.e., a given x) than curve 2. For curve 4, the value of pb =pr is such that Mth ¼ 1 and Me < 1 (as Example 5.1 illustrates), so that pe4 ¼ pb , where pe4 is the exit plane pressure for curve 4 (similar subscripts are used for other curves in Fig. 5.4). A further decrease in pb to a particular pe5 (curve 5) would lead to Me > 1. The particular conditions chosen for curves 4 and 5 are such that the exit plane pressure is equal to the ambient pressure—for curve 4 pe4 ¼ pb is able to adjust and recover to pb because of the subsonic nature in the diverging section. For the case of Me5 > 1, the flow expands (accelerates) so that pe5 ¼ pb ; in the this case, the nozzle is said to be perfectly expanded. For pb ¼ p6 < pe5 , the flow, being supersonic, cannot adjust to pb and expansion continues outside the nozzle (see Example 5.2) and the nozzle is under-expanded (i.e., the gas is not sufficiently expanded at the exit plane to reach pe ¼ pb ). The perfectly expanded condition is achieved by increasing the area ratio Ae =A (for example, by adding further length to the diverging section), or, for a given Ae =A , by increasing pb =pr (say, by reducing pr , or increasing pb , or both as long as the nozzle is still choked). Under-expansion does not occur if the exit flow is subsonic, under which condition the flow can adjust to the ambient condition. Therefore for a given Ae =Ath it is possible to have Mth ¼ 1 (for which Ath ¼ A and either Me < 1 or Me > 1 with isentropic conditions throughout the nozzle) depending on the value of pb . These two possibilities of either subsonic or supersonic exit condition have been numerically illustrated in Example 5.1. These findings help to explain why curves 4 and 5 share a common portion in the converging section and differ in the diverging section. This is because, once the sonic condition is established at the throat, further decreases in pressure in the ambient cannot be felt upstream of the throat (recall the signaling mechanism discussed in Chapter 3). When the condition of Mth ¼ 1 is established, the nozzle is said to be choked. The ratio pb =pr corresponding to curve 4 is called the critical pressure ratio below which Mth ¼ 1. The nozzle flow discussed so far is isentropic and answers the first and second questions at the beginning of this section. The pressure variation in an isentropic nozzle flow can be further explained by using Eqn. (5.8) as follows: (1) For M < 1 throughout the nozzle length, an increase/decrease in the nozzle cross-sectional area, A, causes p to decrease/increase. This result is seen for curves 1–3 in Fig. 5.4. Curve 4 is a special case where Mth ¼ 1 and pb =pr reaches the minimum at throat and recovers in the diverging section due to deceleration to subsonic speeds. For curves 1 through 4, pe ¼ pb due to the subsonic exit flow being able to adjust to the ambient pressure.

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5.3 Standing Normal Shock Wave in Nozzles

141

(2) For M > 1 in the diverging section, Mth ¼ 1 is required and an increase in A causes p to decrease. Curve 5 in Fig. 5.4 is a special case because pb ¼ pe5 and the nozzle is perfectly expanded. For pb ¼ p6 < pe5 the pressure profile remains the same as curve 5 due to the supersonic exit flow’s inability to adjust to the ambient pressure; the flow continues to expand outside the nozzle exit (curve 6). How will the nozzle flow behave with values of pb =pr between curves 4 and 5? In this range of pb =pr values, the flow immediately downstream of the throat is supersonic, but if the flow continues to accelerate/expand toward the nozzle exit, such as that indicated by curve 5, the exit pressure falls below that of the ambient. The adjustment of the exit pressure has to be accomplished through a shock wave due to the supersonic nature of the flow. Let point C in Fig. 5.4 denote the scenario in which supersonic acceleration continues until the flow reaches the exit where the adjustment to pb requires a normal shock at the exit plane so that pb =pr ¼ peC =pr . For pe5 =pr < pb =pr < peC =pr (an example is denoted by curve D in Fig. 5.4), the pressure difference between those of the exiting flow and the ambient is not sufficient for a normal shock at the exit; in this case an oblique shock forms attached to the nozzle lip. The deflection by the oblique shock is to direct the flow inwardly toward the nozzle centerline, as the oblique shock theory of Chapter 4 indicates and as, intuitively, a consequence of the higher ambient pressure squeezing the stream tube downstream of the exit. For pb =pr > peC =pr , the normal shock wave moves upstream of the exit as pb is increased, with curves A and B in Fig. 5.4 as examples (with peA =pr and peB =pr , respectively). Under these back pressures and the resultant normal shock conditions, the flow becomes subsonic in the diverging section downstream of the normal shock and the flow decelerates toward the exit. Because of the subsonic condition, the flow can adjust to the ambient condition so that pe =pr ¼ pb =pr . Further increase in pb above peA eventually moves the normal shock to the throat, where the trivial solution for the shock wave is M1 ¼ M2 ¼ 1 (where subscripts 1 and 2 denote, respectively, states immediately upstream and downstream of the normal shock) and the flow becomes subsonic throughout the entire nozzle (as shown by curve 4 in Fig. 5.4). In summary, the nozzle exit flow can only adjust to the ambient pressure through shock waves for p5 =pr < pb =pr < p4 =pr . Under these conditions, the flow and the nozzle are said to be over-expanded. A normal shock within the nozzle causes a loss in stagnation pressure (i.e., pt2 < pt1 ). Assuming that the flow is everywhere isentropic except across the shock, the downstream stagnation pressure now serves as a reference value for the subsonic decelerating isentropic flow for the remaining diverging portion of the nozzle. For adiabatic walls, the stagnation temperature remains unchanged (Tt2 ¼ Tt1 ¼ constant) throughout the entire nozzle. Does the reference “throat” area, which is also a reference parameter, remain the same as the one upstream of the shock? To answer this question, consider a generic channel where the reference area does not exist in the physical problem, as mentioned in Example 5.2 and now illustrated in Fig. 5.5, where the imaginary extensions and throats are shown. For steady flows, Eqn. (5.13) requires that m_ ¼ constant and pt1 A1 ¼ pt2 A2 . Thus

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142

5 Steady One-Dimensional Flows in Channels Ae* = A2* ( > A2*) A*i = A1* 1

Figure 5.5 Schematic illustrating the imaginary reference cross-sections upstream and downstream of a normal shock. The dashed lines represent the imaginary nozzles.

2

e

i

A2 pt1 ¼ >1 A1 pt2

(5.15)

This result indicates that for the steady state flow, the cross-sectional area of the throat (or the “reference” area) downstream of the normal shock has to be greater than that of the upstream throat. If the channel is without any shock, then A1 ¼ A2 . Implications of this result for supersonic inlet and diffuser designs and the start-up operation of a closed-circuit supersonic wind tunnel will be discussed in the next section. EXAMPLE 5.4 A converging-diverging nozzle is attached to a reservoir containing air at 300 kPa and 300 K. The areas of the throat and the exit are, respectively, 5 cm2 and 10 cm2. (a) What is the range of back pressure for the nozzle to be choked? (b) What is the maximum mass flow rate the nozzle can deliver? (c) What is the range of back pressure for a normal shock wave to exist in the nozzle? (d) What should the back pressure be for a normal shock wave to stand where the cross-sectional area is 7.5 cm2? (e) What is the back pressure for perfect expansion?

Solution – (a) Referring to Fig. 5.4, the choked flow exists for Mth ¼ 1 and pb ≤ pe4 (i.e., either Me < 1 or Me > 1). In this case, Ae =A ¼ ð10 cm2 =5 cm2 Þ ¼ 2 and therefore Me ≈ 0:30 and pe4 =pt ¼ pe4 =pr ¼ 0:9395. For a choked nozzle pb ≤ pe4 ¼ 281:9 kPa; this is the back pressure below which Mth ¼ 1. (b) When the nozzle is choked, any further decrease in pb cannot be sensed by the flow upstream of the throat (e.g., curves 4 and 5 share a common portion in the converging section) and the sonic condition at the throat Mth ¼ 1 gives the maximum mass flow rate. Thus,

Pt A pffiffi γ þ 1 ðγþ1Þ=2ð1γÞ m_ ¼ pffiffiffiffiffiffiffiffi γ 2 RTt   

300  103 Pa 5  104 m2 pffiffiffiffiffiffiffi 2:4 3 kg ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:35 1:4 2 s ð 287 J=kg ⋅ KÞð300 K Þ

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5.4 Supersonic Inlet and Diffuser

143

(c) The upper limit for the back pressure to cause a normal shock in the nozzle is that the shock stands right at the throat and therefore pb ¼ pe4 ¼ 281:9 kPa. The lower limit is pe5 in Fig. 5.4. In this case Ae =A ¼ 2 gives M5 ≈ 2:20, Me5 ≈ 0:5471 and the ratio of the stagnation pressure across the shock For Me5 ≈ 0:5471, pe5 =pt5 ≈ 0:815. Therefore, pt5 =pr ¼ 0:6281. pb ¼ pe5 ¼ ðpe5 =pt5 Þðpt5 =pr Þðpr Þ ¼ 0:815  0:6281  300 kPa ¼ 153:57 kPa. Thus for a normal shock to exist in the nozzle is 153:57 kPa < pb < 281:9 kPa. (d) Let 1 and 2 denote location immediately upstream and downstream of the shock wave. For the shock wave to exist, the flow must be supersonic and must be sonic at the throat. Therefore A1 =A1 ¼ A1 =Ath ¼ 1:5, M1 ≈ 1:86, M ≈ 0:60 and pt2 =pt1 ¼ 0:786 ¼ A1 =A2 . Therefore, Ae =Ae ¼ Ae =A2 ¼  2   Ae =A1 A1 =A2 ¼ ð2Þð0:786Þ ¼ 1:576 and Me ≈ 0:40 and pe =pte ¼ pe =pt2 ¼ 0:896. Hence pb ¼ pe ¼ ðpe =pte Þðpte =pt1 Þðpt1 Þ ¼ ð0:896Þð0:786Þð300 kPaÞ ¼ 211:3 kPa. (e) For perfection, M2 ≈ 2:20 and pb =pr ¼ pe5 =pr ¼ 0:0935. Thus pb ¼ 28:1 kPa. This can also be found using the result in (c) – because M5 ≈ 2:20, pe5 =p5 ¼ 5:480 and therefore without the normal shock at the □ exit and for perfect expansion pb ¼ p5 ¼ 153:57 kPa=5:48 ≈ 28:0 kPa.

5.4 Supersonic Inlet and Diffuser A supersonic diffuser is a converging-diverging nozzle operating in reverse, for the intake of supersonic air streams. The ideal diffuser operation condition is illustrated in Fig. 5.6a, where the free-stream Mach number M∞ is the design Mach number, MD (MD ¼ M∞ ). The design goal is to minimize loss in stagnation pressure (i.e., to maximize the ability of the gases to expand in the downstream nozzle), by achieving isentropic flow throughout the diffuser. Therefore, Ai =AD ¼ Ai =Ath is a function of MD , with Ai being the cross-sectional area of the diffuser inlet and AD , the designed reference area. Operation under the design condition would have the supersonic flow decelerating with no shock wave in the converging section, reaching sonic condition at the throat and then further decelerating in the diverging section to achieve Me < 1 and pe ¼ pb , where pb is now the desired pressure (for example, at the inlet of the engine compressor). Therefore Mth ¼ 1 (the choking condition) and the mass flow rate is limited by the throat cross-sectional area Ath . The following examples help to illustrate the point. From another viewpoint, Mth ¼ 1 is required because Mth > 1 will lead to flow acceleration in the diverging section and Mth < 1 means a normal shock is formed in the converging section. A converging-diverging channel to be used as supersonic diffuser is designed for M∞ ¼ MD ¼ 1:5, p∞ ¼ 10 kPa and p∞ ¼ 216 K. The goal is for the flow to decelerate to subsonic condition and then be fed to an engine with a subsonic speed. What are the inlet and exit cross-sectional areas and the exit

EXAMPLE 5.5

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144

5 Steady One-Dimensional Flows in Channels (a)

Shock

(b)

Spillage Ath = A*

Ath = AD* A*i

M 1 p∞

i Ai

(c)

Mth = 1 (Extremely weak shock at throat)

pb

e

i

Ae

Inlet Ai

Spillage M = 1 th

Ath < AD*

Ath = A*2 1 M Ai ) will not be able to pass through the physical design throat area Ath . The adjustment of the air flow to the mismatch takes place through a normal shock standing at some standoff distance in front of the diffuser inlet, as depicted in Fig. 5.6b. The result is “spillage” of part of the air flow within Ai over the diffuser. The standoff distance is necessary as the subsonic flow downstream of the shock needs a passage for spilling. In Example 5.4 for M∞ ¼ 1:50, part of the air attempting to enter the diffuser will spill – that is the air flow in a cross-sectional area of ðAi  Ai Þ=Ai ≈ 1  1=1:176 ¼ 15% spills, represented by the shaded area in the Fig. 5.6b (exaggerated in size for clarity). During the operation of a supersonic diffuser, there is the so-called start-up problem as the flow regime transitions from subsonic to supersonic. At a moment when M∞ < MD , how would the diffuser function differently from when operated at MD ? For illustration, consider the diffuser geometry in Example 5.5 with and M∞ ¼ 1:40 and MD ¼ 1:50. For M∞ ¼ 1:40, Ai =Ai ≈ 1:115. Therefore, Ai =Ath ¼ ðAi =Ai ÞðAi =Ath Þ ¼ 1:176=1:115 ≈ 1:055, that is, the capture area is larger than the diffuser’s throat area but is still smaller than Ai . This means that only the air flow in the stream tube passing the imaginary cross-sectional area of Ai ≈ 1:055Ath can enter the diffuser and be passed through the throat. The air flow in the area of ðAi  Ai ¼ Ai  Ai ¼ ð1:176  1:115ÞAth ¼ 0:061Ath Þ will spill, and the spillage occurs with a normal shock at a standoff distance in front of the diffuser inlet. Therefore, operating with 1 < M∞ < MD , the spillage is smaller than operating with M∞ ¼ MD . However, it is noted that for M∞ < MD , the mass flow rate through the diffuser is slightly smaller than that for M∞ ¼ MD because even for a larger allowable capture area now Ai ≈ 1:055Ath (a 5.5% increase in the allowable capture pffiffiffiffiffiffiffiffiffiffiffi area), the flow speed (V∞ ¼ M∞ γRT∞ ) is less than that for MD by a factor of 6.7%, corresponding to decrease in M∞ from 1.5 to 1.4.

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5.4 Supersonic Inlet and Diffuser

147

Spillage occurs during the transient operation when M∞ is increased toward MD . The shock wave then moves closer to the diffuser inlet as M∞ ¼ MD is achieved (Fig. 5.6c). Because of the non-zero spillage, Ai < Ai . To eliminate spillage, Ai has to be equal to Ai , then there will be no spillage and the normal shock wave will be standing exactly at the diffuser inlet. Under this condition (shown in Fig. 5.6d), the Mach number downstream of the shock wave (M2 ) is desired to be such that Mth ¼ 1 for the maximum choking mass flow rate allowed by the diffuser. Using numbers provided in Example 5.4, the requirement of Mth ¼ 1 leads to M2 ≈ 0:61 because A2 =A2 ¼ A2 =Ath ≈ 1:176. Therefore the Mach number upstream of the normal shock is M1 ¼ M∞ ≈ 1:81. Thus, avoiding spillage would require M∞ > MD ; that is, over-speeding is needed. The shock wave, however, is not desirable. Since M2 is less than unity under this over-speeding condition, the resultant acceleration in the converging section causes pressure to decrease in the flow direction due to subsonic acceleration from location 2 to the throat. This means that the normal shock wave is not stable and cannot stand still in the converging section. By moving further into the converging section, the shock wave is able to adapt a smaller shock strength, because the Mach number upstream of the new shock location decreases. This new conditions begets yet another new downstream location. Eventually, the shock wave moves past the throat and reestablishes itself in the diverging section, as shown in Fig. 5.6e. As a consequence, the shock wave can be swallowed with sufficient over-speeding. This is because under the M∞ > MD condition, the capture area is smaller than Ai in Fig. 5.6b (i.e., Ath > AD ), so Ath is greater than needed to decelerate to Mth ¼ 1 so that Mth > 1 and the flow in the diverging section is supersonic. The shock wave in the diverging section may be eliminated, depending on the back pressure. However, the captured mass flow rate exceeds that which the diffuser is designed for (again because Ath > AD ). Therefore, once the shock is swallowed, the operation can be returned to the design condition so that Ath ¼ AD again, for which an extremely weak shock (if shock cannot be avoided at all, depending on the back pressure) stands at the throat location. Because the shock is nearly a trivial solution for normal shock problem (M1 ≈ M2 ), the loss in stagnation pressure is nearly zero, which is the desirable result, as illustrated in Fig. 5.6a. The unstable behavior of a normal shock in the converging section is unlike the normal shock wave in the diverging section of a nozzle. In the post-shock (subsonic) region in the diverging section the flow decelerates and the pressure increases in the flow direction. For the shock wave to move downstream, pb has to be decreased, as discussed in Chapter 4. The downstream movement of the normal shock wave increases the Mach number upstream of it and increases the shock strength and the pressure in the subsonic region downstream of it. This creates a contradiction in that in subsonic flows, pressure should equilibrate to that in the ambient. Therefore, it is not possible for a normal shock to move downstream in the diverging section. Over-speeding can be relatively easy to achieve for swallowing the shock wave for moderate values of MD , but becomes increasingly difficult as MD is increased. For example, MD ¼ 1:8 and 2.0, the amount of over-speeding (M∞ ¼ 3:1 and M∞ > 25

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148

5 Steady One-Dimensional Flows in Channels

(a)

Nozzle with enlarged throat

Designed nozzle

(b)

Ath = AD* Mth = 1

M ∞ = MD > 1

M 1

M ∞ = MD > 1

M >1

(d)

M >1

pb

Figure 5.7 Swallowing normal shock by enlarging the throat area – (a) solid and dashed lines represent the design and the enlarged operation conditions, respectively; (b) when starting from rest up to M∞ ¼ MD normal shocks exist upstream and at the inlet (a start-up problem similar to that described in Fig. 5.7); (c) by enlarging the throat area, the normal shock is swallowed; (d) the throat is restored to the design condition.

Ath = AD* Mth = 1

M ∞ = MD > 1

M >1

M 1 and the shock wave moves past the throat, similar to what over-speeding achieves in swallowing the shock wave, as depicted in Fig. 5.7. Once swallowing is achieved, Ath =AD ¼ 1 can be restored operation at MD . Such a scheme for swallowing the shock in the starting phase of supersonic diffuser operation is also used for starting supersonic wind tunnels, as will be discussed in the following section. If a supersonic diffuser is designed for MD ¼ 2:2, during the operation with M∞ ¼ 2:2 how much should the throat open up to swallow the shock wave while maintaining Mth ¼ 1?

EXAMPLE 5.7

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5.5 Continuous Closed-Circuit Supersonic Wind Tunnel

149

Solution – For M∞ ¼ 2:2; Ai =Ai ¼ Ai =Ath ≈ 2:005. Assume that the shock wave in front of the diffuser lip is a normal shock and let subscripts 1 and 2 denote the upstream and the immediately downstream regions of the shock. Then M1 ¼ 2:2 gives 0 M2 ¼ 0:547. Then Mth ¼ 1 requires A2 =A2 ¼ Ai =Ath ≈ 1:26. Therefore,  0 0  Ath =Ath ¼ ðAi =Ath Þ= Ai =Ath ¼ 2:005=1:26 ≈ 1:59. The throat area needs to be □ increased by 59% to swallow the shock wave.

5.5 Continuous Closed-Circuit Supersonic Wind Tunnel For ideal and full pressure recovery in continuous closed-circuit supersonic wind tunnels, double-throated design is preferred (shown in Fig. 5.8). To generate supersonic flow in the test section, a converging-diverging nozzle is necessary, although the diffuser is not. Without the diffuser to properly (or isentropically) slow down the flow exiting the test section, shock waves may occur in the flow passage leading back to the compressor, resulting in loss in stagnation pressure that would require more compressor work to maintain the operation. The deceleration of the supersonic flow using a diffuser is beneficial. An open-circuit (also called blowdown) type supersonic wind tunnel does not “recycle” the gas and therefore has no need for a diffuser. Similar to the supersonic diffuser for aircraft engines, there is also a start-up problem for the double-throated supersonic wind tunnel. During the start-up, the flows in the nozzle and the diffuser are not in reverse to each other. The following discussion of the start-up operation assumes isentropic conditions everywhere in the wind tunnel except when crossing shock waves and receiving work from the compressor to maintain pti ¼ pte . For a designed Mach number in the test section, MD , there is a corresponding value of ATS =Ath1 and Ath1 ¼ Ath2 During the steady state operation, the pressure distribution for an isentropic flow from the inlet to the exit of the double-throated section is shown in Fig. 5.8. This means that the flow conditions in the diffuser are exactly the reverse of those in the nozzle, with Mth1 ¼ Mth2 ¼ 1, pi ¼ pe and pti ¼ pte . At the beginning of the start-up, pi > pb has to be established. As pb =pi becomes sufficiently low to choke the nozzle, the throat conditions in the nozzle become sonic and Mth1 ¼ 1. A further decrease in pb =pi (or pb =pti ) would result in supersonic conditions, possibly with a normal shock wave, in the diverging section of the nozzle (as described in Chapter 4). The existence of a normal shock would cause loss in stagnation pressure, with the maximum loss occurring when it stands at the exit of the nozzle (i.e., at the inlet to the test section). Under these start-up conditions, the second throat area must be larger than the first for the flow to pass through because pt2 A1 Ath1 ¼ ¼ 1)

Test Section

M >1

M Ath;1 .

is no preferred location for the shock to stand within the test section due to the isentropic assumption everywhere else and the associated result that any location produces the same shock loss. To ensure supersonic conditions throughout the test section, the shock has to be moved in the downstream direction and to eventually be swallowed into the diverging section of the diffuser. To accomplish swallowing requires further increase in Ath2 , similar to supersonic inlets described in Section 5.3. Once the swallowing is completed, Ath2 can be returned to that for the design condition again, i.e., Ath2 ¼ Ath1 . An example of how to open the diffuser’s throat during starting phase has been given in Example 5.7.

5.6 One-Dimensional Flow in a Converging Channel Flow in a converging channel can be considered a special case of that in a convergingdiverging channel. Figure 5.9 shows a converging nozzle attached to a reservoir having a pressure pr , with pressure distribution through the nozzle length. These results are similar to those in the converging-diverging nozzle, in that the choking condition is reached pb or pb =pr is sufficiently low. Because the nozzle exit has the narrowest cross-sectional area, it is also the throat area under choked conditions,

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Problems

151

pr

P M

pb = pr Me < 1

pb = pe = pr /( γ+1 ) γ/(γ –1)

Me = 1

2

pb < pe = pr /( γ+1 ) γ/(γ –1)

Me = 1

2 (Expansion outside nozzle)

Figure 5.9 Pressure distribution through the length of a converging nozzle attached to a reservoir having a pressure pr .

that is, Mth ¼ Me ¼ 1 and Ae ¼ Ath ¼ A . Under the choked condition, Me ¼ 1 and pe =pr ¼ ½ðγ þ 1Þ=2γ=ðγþ1Þ (= 0.5283 for γ ¼ 1:4). This means that for any value of pb such that pb =pr < ½ðγ þ 1Þ=2γ=ðγþ1Þ , the mass flow rate does not increase further. Such a result differs from those for the converging-diverging nozzle, where the pressure ratio pb =pr for choking is dependent on Ae =Ath , as Part (a) of Example 5.4 demonstrates. For converging nozzles with pb =pr < ½ðγ þ 1Þ=2γ=ðγþ1Þ , the flow continues to expand outside the nozzle (see Fig. 5.9). The mass flow rate for a choked converging nozzle is thus



Pt A pffiffi γ þ 1 ðγþ1Þ=2ð1γÞ Pr Ae pffiffi γ þ 1 ðγþ1Þ=2ð1γÞ ¼ pffiffiffiffiffiffiffiffiffi γ m_ ¼ pffiffiffiffiffiffiffiffi γ 2 2 RTt RTr

(5.16)

which for a given nozzle and Ae is proportional to pr . It is natural to expect no supersonic flow or shock wave within the nozzle. This is because a converging nozzle connected to a reservoir cannot generate supersonic flows.

Problems Problem 5.1 Derive the differential form of momentum conservation, Eqn. (5.3), using Reynold’s transport theorem, using the approach shown in Chapter 2. Problem 5.2 Repeat the calculations for Example 5.4 with a new exit cross-sectional area Ae ¼ 7:5m2 . Observe and explain the difference between these new results and those of Example 5.3.

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152

5 Steady One-Dimensional Flows in Channels

Problem 5.3 A converging-diverging nozzle is connected to a high pressure air reservoir (at Pr and Tr ) to expand the air to an ambient pressure pb . The exit to throat area ratio Ae =Ath is 3. If a normal shock wave stands at a location where the cross-sectional area is equal to 2Ath , what is the value of pb ? Problem 5.4 A converging-diverging channel as supersonic diffuser is design for M∞ ¼ MD ¼ 2:2, p∞ ¼ 10 kPa and p∞ ¼ 216 K. The goal is for the flow to decelerate to a subsonic condition and then be fed to an engine with a subsonic speed. What is the throat area if the flow is to be isentropic throughout the diffuser? What are the exit cross-sectional area and the flow speed if pe ¼ 90 kPa is desired? Problem 5.5 A converging-diverging nozzle is connected to a reservoir containing air at 500 kPa and 200 °C. The air is to be isentropically expanded to a pressure of 100 kPa at a rate of 15 kg/s. Find the cross-sectional areas where the pressure is 373.2 kPa, 152.8 kPa, and 100.0 kPa. What are the Mach numbers at these locations? Problem 5.6 A converging-diverging nozzle is connected to a reservoir containing air at 1 MPa and 500 K. The nozzle exit has a cross-sectional area of 700 mm2. The flow is isentropic everywhere except across a normal shock standing at a location having a local Mach number of 2.5 and a cross-sectional area of 500 mm2. Find the flowing values: (a) the throat cross-sectional area, (b) the mass flow rate, (c) the exit plane Mach number, velocity, and static pressure, and (d) the entropy change across the shock. Problem 5.7 A converging-diverging nozzle with an exit to throat area ratio (Ae =Ath ) of 2.5 is attached to a high-pressure air reservoir having a pressure pr . Under a certain operation condition, there is a standing normal shock wave at a location having an area ratio A=Ath ¼ 2. What is the back pressure pb ? What should be the back pressure for the nozzle to be free of normal shock waves? Problem 5.8 Find the over-speeding necessary to swallow normal shock waves for diffusers designed for flight Mach numbers of 1.8 and 2.0, respectively. If a variable throat is to swallow the shock, how much opening of the throat is needed for these two design Mach numbers? Problem 5.9 Assume that it is possible to infinitely overspeed. What is the theoretical limit of the design Mach number? Problem 5.10 A blow-down type supersonic wind tunnel uses the ambient air as its reservoir. The supersonic test condition is established by using a vacuum chamber to draw air through the converging-diverging nozzle, followed by the test section, as here shown. The run time is limited by the time when the chamber achieves a pressure that causes a normal shock wave to appear in the test section. Find the expression for the run time in terms of the test section Mach number, MTS , the throat area, Ath , chamber volume, V, and the reservoir condition. If MTS ¼ 2:5 is desired, with the chamber volume equal to 20 m3, the air at 101 kPa and 298 K, the throat area Ath ¼ 1  102 m2 , what is the available time for testing?

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Problems

153

Problem 5.11 During the start-up operation of a supersonic wind tunnel, with the test section to throat area ratio ATS =Ath1 ¼ 4, a normal shock wave appears in the test section as shown. How much does the throat of the diffuser (Ath2 ) have to be opened up, compared to its steady state operation value, to swallow the shock wave? If, at this new value of Ath2 , a normal shock exists at throat 2, what is the pressure downstream of the shock if the test section pressure is 20 kPa? Problem 5.12 A supersonic diffuser is designed for a test section Mach number MTS ¼ 2:5 using air. To swallow the starting shock, how much does the throat crosssectional area have to be increased? Repeat for MTS ¼ 3:5 and make observations by comparing the results of the two design Mach numbers. Problem 5.13 Derive the following expression for the mass flow rate m_ for an unchoked converging nozzle. That is, for an unchoked converging nozzle  γ pr γ þ 1 γ1 < 2 pb vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi " # u 2ðγ1Þ γ1 u 2γ pr γ pr γ 1 t  ⋅ ð pb A e Þ m_ ¼ γ1 RTr pb pb For comparison the choked nozzle mass flow rate is given by Eqn. (5.16) and is a linear function of pr and independent of both pr =pb and pb . Problem 5.14 If a converging-diverging nozzle is designed to expand the flow from a reservoir to vacuum, the velocity is thus the maximum achievable (i.e., Vmax ). Assume an isentropic flow, then show that

Vmax

sffiffiffiffiffiffiffiffiffiffiffi 2 at ! γ1

where at is the speed of sound based on the reservoir temperature. Provide a physical reason why Vmax does not approach infinity. Problem 5.15 A converging nozzle is connected to an air reservoir at 500° C and 300 kPa. At the nozzle exit, the air velocity is 300 m/s. Find the Mach number, temperature, pressure, and density at the exit plane. pffiffiffiffiffiffiffiffiffiffi 0 0 Problem 5.16 Define a Mach number M so that M ≡ V= γRTt . Show that along a streamline in an isentropic flow:

γ1 2 M M ¼ M 1þ 2 0

1=2

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154

5 Steady One-Dimensional Flows in Channels

Then use this expression to find the exit Mach number of Problem 5.15. (This result should help to demonstrate that Mach number based on the stagnation properties can serve as a reference parameter.) Problem 5.17 To simulate a supersonic flight condition with an air speed of 738.0 m/s at an altitude of 15 km, a wind tunnel is designed to operate at the same Mach number and pressure, but with a temperature of 8° C to avoid condensation of water vapor. What should be the ratio of cross-sectional to throat areas? What should be the reservoir pressure and temperature? Assume an isentropic flow from the reservoir exit to the test section. Problem 5.18 Use the thrust force equation for a rocket nozzle, following Reynolds transport theorem to give ⋅

F ¼ m Ve þ Ae ðpe  pa Þ where F, ṁ, pa, Ve, Ae, and pe are the thrust force, mass flow rate, ambient pressure, exit velocity, cross-sectional, and exit pressure, respectively, to show that 

γþ1   12
γþ1 pe γ 2 2 γ1 (1) F ¼ pt A þ Ae ðpe  pa Þ; 1  γ1 γ1 pt (2) the maximum thrust for a given rocket chamber pressure (i.e., the total pressure pt) is achieved with pe=pa and (3) for maximum thrust h Ae ¼ A h

2 γþ1

2 γ1

i

γþ1 2ðγ1Þ

h iγþ1 2γ pa pt

i12 
γ1 γ pt pa

1

12 :

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6

One-Dimensional Flows with Friction

Compressible channel flows may very well experience friction and heat transfer in addition to area changes. Wall friction naturally exists, while heat transfer is likely due to large temperature variations as Mach number varies. A typical example is the nozzle flow in the converging-diverging nozzle of a rocket to where there exist simultaneous effects of area change, friction, and heat transfer. In this chapter the effect of friction is analyzed for steady compressible flows. To concentrate on the effect of friction, one-dimensional flow in constant-area channel with adiabatic walls and without external work is considered. As in previous chapters, an ideal gas having constant specific heats (i.e., a perfect gas) is assumed.

6.1 Fanno Line Flow Consider a steady one-dimensional flow in a channel of constant cross-section area with adiabatic walls and no external work, as shown in Fig. 6.1. Such a flow is called a Fanno line flow. The conservation of mass requires that ρV ¼ constant

(6.1)

As no heat transfer occurs through the adiabatic wall and no external work is done, the energy conservation is 1 ht ¼ cp Tt ¼ h þ V 2 ¼ constant 2

(6.2)

Because of the friction effect, entropy change occurs. The change is given by Eqn. (2.37): ds ¼ cv–

dT d–v þR T –v

When expressed in terms of density change, this expression becomes,

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156

6 One-Dimensional Flows with Friction δQ = 0 δW = 0

M

• pb

Figure 6.1 Steady one-dimensional flow in a channel of constant cross-section area with adiabatic walls and no external work (i.e., a Fanno-line flow).

x

ds ¼ cv–

dT dρ R T ρ

(6.3)

Assuming that the state at location 1 is a reference state and that specific heats remain constant throughout the channel, integration of Eqn. (6.3) yields s  s1 ¼ cv– ln

T ρ  Rln T1 ρ1

By using continuity, Eqn. (6.1), this expression becomes s  s1 ¼ cv– ln

T V þ Rln T1 V1

(6.4)

A T-s diagram is useful in explaining the flow, and it can be obtained by replacing velocity by temperature using the energy equation, Eqn. (6.2): pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (6.5) V ¼ 2ðht  hÞ ¼ 2cv ðTt  T Þ Substituting Eqn. (6.5) into Eqn. (6.4) yields







Δs s  s1 T R Tt  T T γ1 Tt  T ln ¼ ¼ ln ln þ ¼ ln þ þ constant cv– T1 2cv– Tt  T1 T1 2 Tt  T1 cv– (6.6) where relations R ¼ cp  cv– and cp ¼ γcv– are used. By further using these relations, Eqn. (6.6) leads to



Δs s  s1 1 T γ1 Tt  T 1 γ1 ln lnðTt  T Þ þ constant ¼ ¼ ln þ ¼ lnT þ cp γ T1 2γ Tt  T1 γ 2γ cp (6.7) Equation (6.7) provides a relationship for plotting the T-s diagram, which is qualitatively sketched in Fig. 6.2, where the T-s curve is the Fanno line for a given mass flow rate. The horizontal axis could be ðs  s1 Þ=cp ¼ Δs=cp or simply Δs, while the vertical is T. Because the friction effect is irreversible, entropy can only increase when no heat can be taken away from the flow (i.e., Δs ¼ ðht  ht1 Þ=T < 0 is not possible for the adiabatic channel). Therefore, Δs is also qualitatively related to Δx, the distance traveled by the flow, in Fig. 6.2; the longer the distance travels, the larger

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6.1 Fanno Line Flow

157

T M1

Accelerating x, s, Δx, Δs, or Δs /cp

the entropy increase. The quantitative relation between Δs and Δx will be derived in the next section. It can be seen in Fig. 6.2 that there exists a maximum entropy along the T-s curve denoted as Psonic , where dΔs=dT ¼ 0 and M ¼ 1 (and, thus subscript “sonic”). Point Psonic divides the T-s curve into a subsonic branch and a supersonic branch. The sonic condition is established in the following, followed by the discussion of general features of the two branches. Referring to point Psonic in Fig. 6.2, dΔs=dT ¼ 0 is obtained by differentiating both sides of Eqn. (6.7):

1 dΔs 1 1 γ1 1 ¼  ¼0 cp dT γ T 2 Tt  T The energy conservation requires that ðTt  T Þ ¼ V 2 =2cp , which along with cp ¼ γR=ðγ  1Þ can be substituted into the above expression to yield

 1 γ1 1 1 γ1 2 1 γ  1 2 γRT 1    1 1  M2 ¼ 0 ¼ ¼ ¼ 2 2 T 2 Tt  T T 2 cp V T 2 γ1 V T Thus, M ¼ 1and s ¼ smax at point Psonic :

(6.8)

Because the wall is adiabatic, Tt is constant along a given T-s curve. Since at any location in the channel and the corresponding point on the T-s curve (Fig. 6.2), T 1 ¼ γ1 2 Tt 1 þ 2 M It is apparent that the upper branch (with a smaller value of M and, thus, higher temperature T) is the subsonic branch, while the lower branch is supersonic. Because

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158

6 One-Dimensional Flows with Friction

both s and x increase in the flow direction on the T-s diagram, it is concluded (1) that for flows in an adiabatic constant-area channel, a subsonic flow accelerates with increasing M toward the sonic condition, i.e., toward point Psonic on the T-s curve (with increasing Δx and Δs and decreasing T), and (2) that a supersonic flow decelerates with decreasing M toward the sonic condition, i.e., toward point Psonic on the T-s curve (also with increasing Δx and Δs and increasing T). How does this similar qualitative increase in entropy cause very different flow behaviors in subsonic and supersonic flows, with the former accelerating and the latter decelerating? The results from nozzle flow analysis coupled with the development of a boundary layer due to viscosity should help to explain such qualitative differences. Viscosity causes the no-slip boundary condition at the wall, resulting in a boundary layer near the wall in which the effect of the wall is felt by the fluid and the fluid velocity within the boundary layer is smaller than that of fluids at a sufficient distance from the wall. That is, the flow outside the boundary layer does not feel the effect of the wall. Because of slower velocities, the mass flow rate within the cross-section of the boundary layer is smaller than what it would be without the effect of viscosity. A displacement thickness near the wall can be calculated, as frequently done in boundary layer flows (Fox et al., 2004); it represents a near-wall region within which there is no flow (i.e., a dead zone). As the effect of viscosity accumulates in the flow (x) direction, the displacement thickness increases, while steady-state mass flow rate remains the same throughout the channel. For the constant-area channel, this development amounts to a reduction in the effective cross-sectional area, effectively resulting in a converging channel (dA=dx < 0). Thus it becomes clear that a subsonic flow accelerates, while a supersonic flow decelerates (as described in Chapter 4), toward the sonic condition at point Psonic . Because the sonic condition is associated with the maximum entropy on the T-s curve, it is not possible for a subsonic flow to accelerate to become supersonic by adding channel length, as the turning around point Psonic would require a decrease in entropy. Neither is it possible to decelerate a supersonic flow to subsonic conditions for the same reason. For given inlet conditions, the sonic condition can be reached given a sufficiently long channel (or given a sufficient increase in entropy) and a sufficiently low back pressure. This critical length for reaching M ¼ 1 (denoted by L , following the convention of using the superscript * for sonic conditions) depends on the initial Mach number, as can be seen in Fig. 6.2. For the channel length not exceeding L , conditions at any location within the channel can be identified on the T-s curve. But what happens if channel length is longer than L , as shown by L2 > L in Fig. 6.3a? For subsonic flows, the fluid can sense the extra channel length and would adjust accordingly by reducing the mass flow rate (m_ 2 < m_ 1 , as shown in Fig. 6.3b), while maintaining M ¼ 1 at the end of the channel. The reduction in m_ can be understood by considering that the displacement thickness has increased due to additional channel length, further reducing the effective cross-sectional area at the sonic location. Furthermore, the stagnation pressure becomes lower with the extra length (this

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6.1 Fanno Line Flow (a)

159

δQ = 0 δW = 0

Inlet

M=1

m• 1

Mi1= 1

pb

L*1 = L1 L2 > L1= L1*

Mi2< Mi1

m• 2 < m• 1

M=1

L2= L2* (b) T

• • m 2 ( L , A2 < A1 and Pt2 < Pt1 . The mass flow rates equation, Eqn. (5.16), can be used for comparing m_ 1 and m_ 2 :

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160

6 One-Dimensional Flows with Friction





Pt2 A pffiffi γ þ 1 ðγþ1Þ=2ð1γÞ Pt1 A1 pffiffi γ þ 1 ðγþ1Þ=2ð1γÞ ṁ2 ¼ pffiffiffiffiffiffiffiffi2 γ < pffiffiffiffiffiffiffiffi γ ¼ ṁ1 2 2 RTt RTt

(6.9)

In this case, the channel is choked, meaning that further lowering pb does not lead to _ although increasing the channel length results in further decrease, an increase in m, _ Also in this case L2 becomes the new critical length (L2 ¼ L2 ) for the new in m. Fanno line of m_ 2 . A supersonic flow cannot sense the fact that L2 > L . It adjusts to the extra _ The flow downlength by means of a normal shock while maintaining the same m. stream of the normal shock can sense the remaining length and can adjust to the back pressure because it is subsonic. As a consequence, the location of the shock wave has to be dependent on the back pressure (pb ), because the location needs to be such that the adjustment pe ¼ pb is achieved. The above analysis is based on continuity and conservation of energy, without invoking the momentum equation. It is apparent that friction force affects the flow speed and leads to the formation of the boundary layer and the displacement thickness that reduces the exit cross-sectional area. Without explicitly considering viscous effects, the result of Eqn. (6.9) is only qualitative, as stated, and so is the discussion of the characteristic of the T-s diagram. Questions remain regarding (1) how to determine the critical length, L , (2) the flow conditions throughout the channel, including those at point Psonic that serve as the reference conditions for a given Fanno line, (3) the location of the normal shock if it exists. The following section describes derivations leading to answers to these questions.

6.2 Equations of Fanno Line Flow and Critical Length L* Inferring from the results of Fig. 6.2 and the previous discussion, the value of L should be dependent on the Mach number at the channel inlet, whether the inlet is subsonic or supersonic. To determine L as a function of M, first consider the momentum conservation for the differential control volume in Fig. 6.4. While the continuity is given by Eqn. (6.1), the momentum equation can be derived by applying the Reynolds transport theorem, Eqn. (2.5):

δQ = 0 δW = 0 p

ρ

T V M

p + dp ρ + dρ T + dT V + dV M + dM

Figure 6.4 Control volume for Fanno line flow analysis.

x

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6.2 Equations of Fanno Line Flow and Critical Length L*

X

~ F i i

 system

¼

∂ ∂t

ð

~ d–V þ ρV

C– V

ð

161


 ~ ~ ρV ~ ⋅ dA V CS

Under the steady-state assumption, this expression is reduced to pA  ð p þ dpÞA  τ w Aw ¼ ½ðρ þ dρÞðV þ dV Þð AÞðV þ dV Þ  ðρVAÞV ¼ ρVAdV where the continuity ðρ þ dρÞðV þ dV Þ ¼ ρV is used, and τ w and Aw are the wall shear stress and the wall surface area, respectively. This expression is further reduced to Adp  τw Aw ¼ ρVAdV

(6.10)

A hydraulic diameter, Dh , can be defined to express Aw in terms of A so that Eqn. (6.10) can be further simplified: Dh ≡

4  ðCross-sectional areaÞ 4 A ¼ wetted perimeter P

(6.11)

It can be easily shown that Dh ¼ D for a circular duct with a diameter D, Dh ¼ W for a square with a width W on each side, and Dh ¼ 2 WH=ðW þ H Þ for a rectangular channel with W and H as its width and height, respectively. Therefore, Aw ¼ Pdx ¼

4 Adx Dh

and Eqn. (6.11) becomes Adp  τ w

4 Adx ¼ ρVAdV Dh

(6.12)

Define a friction coefficient (also called the Darcy friction coefficient), f, as f≡

4τw 1 2 2 ρV

Equation (6.12) becomes 1 dx dp  ρV 2 f ¼ ρVdV 2 Dh

(6.13)

Dividing terms in the above expression through, by either p or ρRT (¼ ργRT=γ ¼ ρa2 =γ) yields dp 1 dx dV þ γM2 f ¼0 þ γM2 p 2 Dh V

(6.14)

To express L as a function of a Mach number will require that dp=p and dV=V in Eqn. (6.14) be also expressed in terms of a Mach number before integration can be carried out. The ideal gas law can be written as

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162

6 One-Dimensional Flows with Friction

lnp ¼ RTlnρ þ ρTlnR þ ρRlnT Differentiating this expression and then dividing the resultant terms on the left-hand side by p and those on the right-hand side by ρRT yields dp dρ dT ¼ þ p ρ T

(6.15)

The continuity equation, ρV ¼ constant; can be similarly manipulated to yield dρ dV ¼ ρ V

(6.16)

Substituting Eqn. (6.16) into Eqn. (6.15) and then the result for dp=p into Eqn. (6.14) leads to  dV dT 1 dx  þ γM2 f ¼0  1  γM2 T 2 Dh V

(6.17)

Equation (4.8) indicates

γ1 2 T 1þ M ¼ Tt 2

(4.8)

and Tt ¼ constant for the Fanno line flow. Differentiation of Eqn. (4.6c) produces
 γ1 2 d 1 þ M 2 dT ¼ (6.18) γ1 T 1þ M2 2

pffiffiffiffiffiffiffiffiffi Since V ¼ M γRT , similar manipulation results in dV dM 1 dT ¼ þ V M 2 T Substituting the expressions for dT=T and dV=V into Eqn. (6.17) yields  2
3 γ1

2 dx γ þ 1 4d 1 þ 2 M 2 MdM5 2 dM f þ ¼  Dh 2γ M2 γ M3 1 þ γ1 M2

(6.19)

(6.20)

2

Integrating Eqn. (6.20) over the length of L for the range of M from M to 1 yields !

γþ1 2 fL γþ1 M2  1 2 M ¼ (6.21a)  ln 2γ γM2 Dh 1 þ γ1 M2 2

Values of fL =Dh can be easily calculated for given values of M, which can be subsonic or supersonic, but not vice versa. For convenience, values of fL =Dh are tabulated against M in Table B.3. For M ¼ 1, fL =Dh ¼ 0. For M ! ∞,



  fL γþ1 γþ1 1 ¼ ln  2γ γ1 γ Dh M!∞

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6.3 Reference State in Fanno Line Flow

163

For γ ¼ 1:4, this limiting value is 0.822. Thus, for the supersonic branch of the Fanno line, fL =Dh falls in the range from 0 to 0.822. For the subsonic branch, fL =Dh can be large as M ! 0

  



fL γþ1 γþ1 2 1 M ¼ ln þ 2γ 2 γM2 M!0 Dh M!0 By applying the l’Hospital rule to the ratio of the two terms on the right-hand side of this expression results in fL =Dh ! ∞ as M ! 0. It is clear that L is a function of M, and therefore serves as the reference length over which the flow with M would have to travel to reach the sonic state, point Psonic on the T-s (or T-x) curve shown in Fig. 6.2. A distance short of L would indicate that the flow is still subsonic or supersonic. The following two examples demonstrate how to determine L and then use it to determine M if the distance traveled is L < L . EXAMPLE 6.1 Air flow enters an insulated circular duct with a Mach number of 0.3, and the duct has a diameter of 5 cm and a friction coefficient of 0.02. What is the length of the duct needed for the flow to reach the sonic condition at the exit of the duct? What is the Mach number if the duct length is 1 m?

Solution – For M1 ¼ 0:3 and γ ¼ 1:4, fL1 =Dh ¼ 5:2993. Therefore, cm ¼ 1; 324:8 cm ¼ 1:325 m; for the sonic exit, L1 ¼ 5:2993  Df h ¼ 5:2993  50:02  L1 ¼ L1 : fL

fL

0:021000 1 ¼ 1:2993. Thus, Me ≈ 0:475. For Dh2 ¼ Dh1  fL 5 Dh ¼ 5:2993  Thus by traveling a distance of 1 m, the M1 ¼ 0:3 flow does not reach the sonic condition. The M ≈ 0:475 would need a length of 0.325 m, shorter than 1.325 m □ for the M ¼ 0:3 flow, to reach Me ¼ 1.

EXAMPLE 6.2 Air flow enters an insulated circular duct with a Mach number of 2.5, and the duct has a diameter of 3 cm and a friction coefficient of 0.02. What is the length of the duct needed for the flow to reach the sonic condition at the exit of the duct? What is the exit Mach number if the duct length is 50 cm?

Solution – For M1 ¼ 2:5 and γ ¼ 1:4, fL1 =Dh ¼ 0:432. Therefore, cm ¼ 64:8 cm; for the sonic exit, L1 ¼ L1 : L1 ¼ 0:432  Df h ¼ 0:432  30:02

For

fL2 Dh

¼

fL1 Dh

0:0250 1  fL ¼ 0:099. Thus, Me ≈ 1:40. 3 Dh ¼ 0:432 

The flow does not travel a sufficiently long distance to reach the sonic state.

□

6.3 Reference State in Fanno Line Flow The flow that has traveled the length L thus reaches sonic conditions. Due to the non-isentropic process, there is loss in stagnation pressure. Therefore pt does not remain constant in Fanno line flows and cannot be a reference parameter. Because of

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164

6 One-Dimensional Flows with Friction

the adiabatic wall assumption, Tt remains constant and can serve as a reference parameter. It is desirable to find out the values of pt , p, and T as a functions of a Mach number. Since Tt ¼ constant and T  ¼ Tt =½ðγ þ 1Þ=2 is also a constant, temperature at the sonic condition can be a choice for reference. Following similar reasoning, values of pt and p at M ¼ 1 might also serve as references. Replacing dV=V in Eqn. (6.14) with Eqn. (6.19) leads to dV dM 1 dT ¼ þ V M 2 T Substituting this expression along with Eqn. (6.18) for dT=T into Eqn. (6.14) yields dp 1 dx dM 1 γM2 ðγ  1ÞM2 dM þ γM2 f  ¼0 þ γM2 p 2 Dh M 2 1 þ γ1 M2 M

(6.22)

2

The second term of Eqn. (6.22) can be expressed in terms of Mach number by rewriting Eqn. (6.20) as " #  dx 1  M2 2 dM f (6.23) ¼ Dh 1 þ γ1 M2 γM2 M 2

Substituting Eqn. (6.23) into Eqn. (6.22) and collecting terms yields " # " # γ1 2 2 1 þ γ1 M M dp 1 þ ðγ  1ÞM2 dM dM 2 ¼ ¼ þ 2 γ1 γ1 γ1 2 2 2 p M 1þ 2 M 1þ 2 M 1þ 2 M M
 γ1 2 dM 1 d 1 þ 2 M ¼  M 2 1 þ γ1 M2 2 Integration of this expression from M ¼ M to M ¼ 1 leads to
 2 !1=2 3 γ1





2 ð p 1 1 1 d 1þ 2 M 1 ð γ þ 1 Þ=2 5 ln  ¼ ln ¼ ln4 þ p M 2 M 1 þ γ1 M2 M 1 þ γ1 M2 2

(6.24)

(6.25)

2

Therefore, p ¼ p



1 M

"

ðγ þ 1Þ=2

#1=2

2 1 þ γ1 2 M

(6.26)

where p=p ¼ 1 for M ¼ 1, as expected. Because at a given location, p 1 ¼
γ=ðγ1Þ pt 2 1 þ γ1 M 2 one obtains

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6.3 Reference State in Fanno Line Flow

165

104 103 pt /pt*

102 fL*/Dh

101

p/p* pt /pt*

100

T / T t*

ρ/ρ*

ρ/ρ*

10–1

T / T t*

10–2

p/p* fL*/Dh

10–3 10–4 0.01

0.1

1

10

M

Figure 6.5 Properties of Fanno line flow as a function of M for γ ¼ 1:4.


γ=ðγ1Þ γ1 

 γþ1 2 1 þ M 2 pt p 1 2 γ  1 2 2ðγ1Þ 1þ M ¼  ¼
γ=ðγ1Þ p M γþ1 2 pt γþ1

(6.27)

2


 2 By usingTt ¼ T 1 þ γ1 ¼ constant, 2 M T ðγ þ 1Þ=2 ¼ (6.28) T  1 þ γ1 M2 2 pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi The continuity equation combined with V ¼ γRT and V  ¼ γRT  leads to ρ V ¼ ¼ ρ V



1 M





 2 ðγ  1Þ 2 1=2 1þ M γþ1 2

(6.29)

As expected in Eqns. (6.27) through (6–29), pt =pt ¼ 1, T=T  ¼ 1 and ρ=ρ ¼ 1 at M ¼ 1. The numerical results of these equations are tabulated along with fL =Dh in Appendix F. For γ ¼ 1:4, they are shown graphically in Fig. 6.5. As with flow across a normal shock, the Fanno line flow is adiabatic and onedimensional. The equation for entropy increase across the normal shock is applicable to the Fanno line flows. Thus, s2  s1 pt2 ¼ ln R pt1

(4.19c)

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166

6 One-Dimensional Flows with Friction

Following Example 6.1, if p1 ¼ 150 kPa, T1 ¼ 300 K, L1 ¼ 10 m, find the pressure, temperature, speed, and pressure loss at the end of the duct (location 2 or exit).

EXAMPLE 6.3

Solution – M1 ¼ 0:3; fL2 Dh

¼

fL1 Dh

p1 T1 p1 ¼ 3:619;  ¼ 1:179; ¼ 0:9395; pt1 ¼ 156:7 kPa p T pt1

0:021000 1  fL ¼ 1:2993. Thus, M2 ¼ Me ≈ 0:475 5 Dh ¼ 5:2993 

p2 p2 p 1  150 kPa ¼ 93:5 kPa ≈ 2:257; p ¼   p1 ¼ 2:257  2   3:619 p p p1 T2 T2 T  1  300 K ¼ 292:1 K ¼ 1:148; T ¼   T1 ¼ 1:148  2   1:179 T T T1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi m γRT2 ¼ 0:475  1:4  287  292:1 ¼ 162:7 s pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi m V1 ¼ M1 γRT2 ¼ 0:3  1:4  287  300 ¼ 104:2 s (→ acceleration from location 1 to location 2) V2 ¼ M2

pt1 p1 pt2 ¼ 2:035; ¼ 0:9395 ðlocally isentropicÞ;  ¼ 1:391 pt pt1 pt pt2 ¼

pt2 pt pt1 1 1   150 kPa ¼ 109:1 kPa p1 ¼ 1:391   2:035 0:9395 pt pt1 p1

ðpt2  pt1 Þ ¼ pt2 

pt1 1  150 ¼ 50:6 kPa  p1 ¼ 109:1  0:9395 p1

□

6.4 Fanno Line Flow in a Nozzle Followed by a Constant-Area Duct In practice, the flow in a constant-area duct may be fed by a converging- or a converging-diverging nozzle attached to a reservoir. First consider the case with a converging nozzle followed by a constant-area duct having a length L, as depicted in Fig. 6.6. For convenience, the flow in the converging nozzle is assumed to be isentropic. It is clear that nowhere through the nozzle-duct system can the flow become supersonic for two reasons: First, the flow can at most be sonic at the entrance/inlet to the duct, so Mi < 1. Second, the sonic entrance flow to the duct cannot be accelerated to be supersonic without heat transfer away from the fluid to decrease in entropy for the transition, to turn around the point Psonic in the Fanno line, from the upper, subsonic branch to the lower, supersonic branch shown in Fig. 6.2. Therefore, the flow in the constant-area duct shown in Fig. 6.6 can at most be

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6.4 Fanno Line Flow in a Nozzle i 1 M = M: Pb, Me = 1 (Expansion outside the duct exit) x

Figure 6.6 Fanno-line flow in a converging nozzle followed by a constant-area duct. For pb ≤ pð3Þ the sonic exit condition is established at the duct exit and the flow is frictionally _ does not increase for pb < pð3Þ and the flow continues to choked, that is, the mass flow rate, m, expand outside the duct. For pb ≥ pð4Þ , m_ increases as pb decreases.

sonic and the only possible sonic location is the exit of the duct (i.e., Me ¼ 1). Associated with this scenario of sonic exit is pe ¼ p . If Me < 1 (an example is represented by Curve 1 in Fig. 6.6), the entire flow throughout the nozzle-duct assembly is subsonic and pb ¼ pe , but pb ¼ pe > p and Me < 1. If pb is reduced correspondingly to Curve 2 in Fig. 6.6, an increase in both Me and m_ result, still with pb ¼ pe > p and Me < 1. For conditions represented by Curves 1 and 2, Me < 1. For these two cases, L < L . Even lower pb conditions of Curve 3 in Fig. 6.6 are achieved such that Me ¼ 1, pb ¼ pe ¼ p , and pte ¼ pt . In this case L ¼ L . Further lowering pb would not cause the flow to become supersonic at exit, as shown by Curve 4 in Fig. 6.6. For Curve 4, Me ¼ 1, pb < pe ¼ p , pte ¼ pt , and L ¼ L and the exit flow cannot sense further decreases in the back pressure; thus pb < pe ¼ p and the flow continues to expand outside the duct. Because L ¼ L , Mi remains the same as that represented by Curve 3, the value of m_ for Curve 4 is thus the same as that for Curve 3. Therefore the duct is choked. In summary, pb ≤ pe ¼ p , pte ¼ pt , and L ¼ L for Me ¼ 1. Once the

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168

6 One-Dimensional Flows with Friction

Me ¼ 1 condition is reached, the duct is choked by friction. That is, further decreases in pb do not lead to larger mass flow rates. For a back pressure of Curve 4, one can afford to add duct length and still achieve Me ¼ 1. This would, however, cause an increase in L , as the new physical _ duct length L is L . As a consequence, Mi is reduced and so is m.  If pb ¼ 0, then Me ¼ 1 and pe ¼ p are guaranteed regardless the length of the duct. Then whatever the duct length is, L ¼ L . In this case, a larger L leads to _ For L ¼ L ! ∞, Mi ! 0 (and a smaller M at the duct inlet and thus a smaller m. m_ ¼ 0), as discussed in Section 6.2; this result is expected because no flow can overcome friction over an infinite distance. A constant-area duct is connected to an air reservoir (at 150 kPa and 300 K) through a converging nozzle. At the junction of the nozzle and the duct, the diameter is 5 cm. The friction coefficient of the duct wall is 0.02, and the duct length is 1 m. What is the maximum back pressure for Me ¼ 1, what is the maximum mass flow rate, and what is the pressure due to the frictional duct?

EXAMPLE 6.4

Solution – For Me ¼ 1, pb ≤ pe ¼ p . Let the location of the nozzle-duct junction be designated as “1.” Using the table in Appendix F, pt1 T1 T  ≈ 1:174, and p ≈ 1:970.

fL1 Dh

¼ 0:021000 ¼ 4, M1 ≈ 0:33, 5

p1 p

≈ 3:287,

t

To find p , the duct inlet pressure p1 is needed. For M1 ≈ 0:33, the  isentropic relationship provides ppt11 ¼ pp1r ≈ 0:922. Therefore, p ¼ pp1  pp1r  pr 1 ¼ 3:287  0:922  150 kPa ¼ 42:08 kPa. Therefore, pb ≤ 42:08 kPa for Me ¼ 1. pffiffiffiffiffiffiffiffiffiffiffi p γRT So, T  is needed. Following similar process of m_ ¼ ρ AV  ¼ RT  A  finding p , T ¼

T  T1 1  0:979  300 K ¼ 250:2 K   Tr ¼ 1:174 T1 Tr

Therefore, m_ ¼

 ipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffiffiffiffiffiffiffiffiffi 42:08  103 hπ  2 2 5  10  ¼ A γRT 1:4  287  250:2 4 RT  287  250:2 ¼ 0:365 kg=s

or pffiffiffiffiffiffiffiffiffiffiffi p1 AM1 γRT1 RT1 3 h pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 i 150  10  0:922 π  5  102 ¼  0:33  1:4  287  300  0:979 287  300  0:979 4 m_ ¼

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6.4 Fanno Line Flow in a Nozzle

169

¼ 0:365 kg=s Because pt ¼ pte andpt1 ¼ pr pte p ðpte  pt1 Þ ¼   t  pt1  pt1 ¼ pt pt1





1 1  1 pr ¼  1  150 kPa 1 1:97 1:97

¼ 73:86 kPa

□

Repeat the calculation for the mass flow rate and pressure loss in Example 6.4 except now pb ¼ 0 and the duct length is 3.25 m. Discuss the differences in results.

EXAMPLE 6.5

Solution – Because pb ¼ 0 < p for any realistic value of p , so the duct is choked and Me ¼ 1. fL ¼ 13, M1 ≈ 0:21, According to the table in Appendix F, Dh1 ¼ 0:023250 5 p1 pt1 T1 ≈ 5:205, ≈ 1:189, and ≈ 2:836.    p T p t

p ¼

p p1 1  0:970  150 kPa ¼ 27:95 kPa   pr ¼ 5:205 p1 pr

T ¼

m_ ¼



T  T1 1  0:991  300 K ¼ 250:0 K   Tr ¼ 1:189 T1 Tr

 ipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffiffiffiffiffiffiffiffiffi 27:95  103 hπ  2 2 5  10  ¼ A γRT 1:4  287  250:0 4 RT  287  250:0 ¼ 0:242 kg=s pt ¼ pte ; ðpte  pt1 Þ ¼ ¼





1 1  1 pt1 ¼  1 pr 2:836 2:836

1  1  150 kPa ¼ 97:11 kPa 2:836

Discussion – The mass flow rate is reduced from that in Example 6.4 due to the added duct length. This trend is consistent with the result of the T-s diagram shown in Fig. 6.2; that is, when the duct is already choked (as in Example 6.4 also for pb ¼ 0) further addition of the duct length should lead to a reduced mass flow rate and a new T-s curve. This reduction is accompanied by an increase in the pt2 1 loss in stagnation pressure, which according to s2 s R ¼ ln pt1 also leads to

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170

6 One-Dimensional Flows with Friction L < L*

Throat pr Mi>1

Tr

p pb = 1

pb

pe = pb = pr

(h’) pe = pb, Me < 1, Mth < 1 (h) pe = pb, Me < 1, Mth < 1 (g) pe = pb, Me < 1, throughout duct (f) pe = pb, normal shock in nozzle, Me < 1 Mth =1

(e) pe = pb, normal shock at junction, Me < 1 (d) pe = pb, normal shock inside duct, Me < 1 (c) pe = pb, normal shock at exit, Me < 1 (b) pe = pb, Me = 1

(a) pe > pb, expansion wave outside duct, Me = 1

(a)–(g) Mth = 1: nozzle choked x

Figure 6.7 Fanno line flow in a converging-diverging nozzle followed by a constant-area duct with L < L . Unlike in Fig. 6.5, supersonic flow can exist in the duct for pb ≤ pe as no normal shock exists in the diverging section of the nozzle (pe is the ambient pressure corresponding to Curve e). For pb ≤ pg , the nozzle-duct assembly is choked and m_ does not change. For pb ≤ pðbÞ , the exit Mach number Me ¼ 1.

a higher increase in entropy at the duct exit. As a result the new T-s curve is to □ the right of the one with shorter duct length, as clearly shown in Fig. 6.2. The case of a constant-area duct connected to a reservoir through a convergingdiverging nozzle is shown in Figs. 6.7 (for L < L ) and 6.8 (for L > L ). The flow in the converging-diverging section is again assumed to be isentropic, while the effect of wall friction is present in the constant-area duct. For pb ¼ pr there is no flow and the pressure is the same and equal to pb throughout the entire channel. As the back pressure is lowered (pb < pr ), a flow is induced. For the cases where subsonic flow enters the duct (Mi < 1), the flow would

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6.4 Fanno Line Flow in a Nozzle Throat pr

171

L > L* M >1

M pb, Me = 1

Choked nozzle

Figure 6.8 Fanno line flow in a converging-diverging nozzle followed by a constant-area duct with L > L . Supersonic flow can exist in the duct; however, unlike in Fig. 6.9, normal shock always exists even if the ambient is vacuum. Only for pb ≤ pðbÞ is the exit Mach number Me ¼ 1.

remain subsonic throughout the flow, and at most become sonic at the duct exit (Me ¼ 1). Such a scenario is similar to that of a converging nozzle-duct assembly already discussed. Thus, the case of supersonic flow at the inlet of the duct (i.e., at the nozzle exit) is more relevant for gaining new insight. For the flow to be supersonic at the nozzle exit, the throat Mach number must be unity, that is, Mth ¼ 1. As a consequence, the flow does not sense any further lowering of pb and the mass flow rate reaches the maximum allowed:

Pr A pffiffi γ þ 1 ðγþ1Þ=2ð1γÞ m_ ¼ pffiffiffiffiffiffiffiffiffi γ 2 RTr

(5.13)

where A ¼ Ath . In this case the nozzle, rather than the duct, is choked and Mi > 1. Three scenarios of exit condition are possible: (1) L < L , Me > 1, and pe ≥ pb ; (2) L < L and if the flow at the duct exit has to adjust to pb , the adjustment has to be done through a shock wave and the shock wave location is determined by pb so that pe ¼ pb ; and (3) that if L > L , there would always be a shock wave in the duct, as

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172

6 One-Dimensional Flows with Friction

discussed earlier for the characteristics of the T-s diagram of Fig. 6.2, and the shock wave location is determined by pb so that pe ¼ pb . For the first scenario (L < L , Me > 1, and pe ≥ pb ) for the given Mi , the pressure variation in the nozzle-duct assembly is shown by Curves (a) and (b) in Fig. 6.7. Curve (a) represents the case of pe > pb , while Curve (b) represents the special condition of pb ¼ pe , with Me > 1 for both cases. For back pressure of Curve (a), the flow continues to expand outside the duct. The flow does not have to (and cannot because Me > 1) adjust to pb . The second scenario requires an increase in pb above conditions represented by Curve (b). Such an increase in pb would lead to oblique shocks at the duct exit (i.e., conditions between Curves [b] and [c]) until the back pressure is sufficiently high to cause a normal shock at the exit (Curve [c]). Downstream of the shock, the exit pressure adjusts to the ambient value, that is, pe ¼ pb . Further increases in pb above that of Curve (c) result in moving the normal shock upstream of the duct exit, with Curve (d) as an example. For sufficiently high values of pb , the normal shock moves into the diverging section of the nozzle and the entire flow in the duct is subsonic. The back pressure for Curve (e) causes the normal shock to stand right at the junction of the nozzle and the duct, while pb for Curve (f) moves the shock inside the diverging section of the nozzle. For Curves (f), (g), and (h), Mi < 1, enabling the adjustment for pe ¼ pb . In summary, for Mi > 1 and L < L , the location of normal shock, if it occurs, in the nozzle-duct assembly depends on the back pressure. If the back pressure is sufficiently low, no normal shock exists at all. This conclusion is similar to the existence and the location of a normal shock wave in a choked convergingdiverging frictionless nozzle, as discussed in chapter in Chapter 5. A constant-area duct is connected to a converging-diverging nozzle, which is in turn connected to a reservoir of air at 400 kPa and 400 K. The area ratio of the nozzle is 2.0 while the length to diameter ratio of the duct is 5 and the friction coefficient is 0.02. What is the range of back pressure for a normal shock to exist in the duct?

EXAMPLE 6.6

1

i

2

e

Pr Tr

Shock at inlet 1 i

Shock at exit

M1 = 2.20 Mi = 0.547

M2 = 1.86 Me = 0.60

2

e

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6.4 Fanno Line Flow in a Nozzle

173

Solution – For a normal shock to exist in the duct implies that the Mach number at the A1 1 nozzle exit must be greater than one. As shown in Fig. E6.6, A A ¼ Ath ¼ 2. Therefore, M1 ¼ Mi ¼ 2:20; for which fL1 Dh

¼ 0:02  5 ¼ 0:10
fL Dh ¼ 0:10 and

pi p

≈ 1:95.

1  fL Dh ¼ 0:630. Thus Me ¼ M2 ≈ 0:565.

pe p pi p1 pt1      pr p pi p1 pt1 pr 1  5:341  0:0935  1  400 kPa ¼ 192:58 kPa 1:95

Therefore, for the range of back pressure 47:72 kPa < pb < 192:58 kPa, a normal shock exists in the duct, with the higher and lower limits for it to □ stand at the duct inlet and at the duct exit, respectively. In the third scenario, with Mi > 1 and L > L , a normal shock exists in the nozzle-duct assembly. Figure 6.8 shows the pressure variation in the nozzle-duct assembly as the back pressure is varied. For extremely low back pressures (such as a vacuum), the flow accelerates downstream of the normal shock with decreasing pressure. Because of the low back pressure, the flow will expand and achieve sonic condition at the exit. Note that even for a vacuum back pressure, the subsonic flow downstream of the shock cannot become supersonic, as discussed earlier. The sonic exit condition is still possible with pb for Curves (a) and (b). Curve (b) is a special case with Me ¼ 1 and pe ¼ pb while Curve (a) has Me ¼ 1 and pe > pb . For back pressures below curve (b), the location of normal shock does not change to a further downstream location, because the sonic flow at the duct exit prevents the flow from sensing any lower back pressure and adjusts accordingly. For back pressure higher than that for Curve (b), the normal shock wave moves upstream toward the nozzle, as depicted by Curve (c). For sufficiently higher pb , the normal shock wave moves into the diverging section of the nozzle, as shown by Curve (d), and the entire flow in the duct is subsonic. For pb corresponding to that of Curve (e), the shock wave disappears with Mth ¼ 1, i.e., a choked nozzle. For pb below that for Curve (e), no

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174

6 One-Dimensional Flows with Friction

T

M L*

· (m reduction)

Figure 6.9 The transition from supersonic to subsonic conditions due to shock wave on a given Fanno line (i.e., maintaining the same mass flow rate as the nozzle is already choked). Note that the shock within the duct can be caused by either L > L or by sufficiently high back pressure with L < L . Also shown is the mass reduction due to duct lengthening for the subsonic Fanno line.

M=1

M>1 x, Δx, s, Δs, Δs/cp

more increase in the mass flow rate is possible, as the flow in the nozzle cannot sense such a condition. For pb above Curve (b), the exit Mach number is less than unity, Me < 1 and pe ¼ pb . Figure 6.9 shows the transition from supersonic to subsonic conditions due to the shock wave on a given Fanno line, that is, maintaining the same mass flow rate as the nozzle is already choked. It is noted that the transition is in the direction of increased entropy and the subsonic flow accelerates toward point Psonic . Depending on the value of pb , the exit flow may reach the sonic condition as discussed above. For conditions given in Example 6.6, except that the length to diameter ratio is now 25, find the pressure range for a normal shock to exist in the duct. EXAMPLE 6.7

Solution – fL1 1 Because fL Dh ¼ 0:02  25 ¼ 0:50 > Dh ¼ 0:3609, there will be a normal shock in the nozzle-duct assembly even if the back pressure is a vacuum (in fact this is the case for back pressure below that corresponds to Curve (b) in Fig. 6.8). The question first of all becomes: at what back pressure does the shock stands at the duct inlet? A1 A1 A ¼ Ath ¼ 2. Therefore, the Mach number at the nozzle exit is M1 ¼ 2:20, for p1 which pt1 ¼ pp1r ¼ 0:0935 at the nozzle exit and pp1i ¼ 5:80 across the shock. After fL

the shock the Mach number at the duct inlet is Mi ¼ 0:5471 for which Dhi ¼ 0:747 and

pi p

≈ 1:95. Therefore,

fLe Dh

¼

fLi Dh

fL

 Dh1 ¼ 0:747  0:50 ¼ 0:247, and Me ≈ 0:68,

for which ppe ≈ 1:541.

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Problems

pb ¼ pe ¼

175

pe p pi p1 1  5:480  0:0953  400 kPa     pr ¼ 1:541  1:95 p pi p1 pr ¼ 165:1 kPa:

Therefore, for a normal shock wave to exist in the duct, 0 < pb < 165:1 kPa. Comments – This upper limit of back pressure is higher than that in Example 6.6, simply because for the same shock location at the nozzle-duct junction, there is a longer duct now for the subsonic duct for to accelerate, resulting in a lower exit-plane pressure and, thus, a lower back pressure for the normal shock to stand at the duct inlet. Also differing from Example 6.6 is that the current example has a long □ duct to guarantee a normal shock within the duct.

Problems Problem 6.1 Show the steps leading to Eqns. (6.21) and (6.23). Problem 6.2 A perfectly insulated circular pipe (2.5 m in length and 5 cm in diameter; friction coefficient is 0.02) is connected to an air reservoir (at 700 kPa and 450 K) through a converging nozzle. What is the maximum flow rate that can be delivered by this nozzle-pipe assembly? What is the exit velocity under this condition? Under what back pressure can this maximum mass flow rate be delivered? Problem 6.3 Referring to Problem 6.2, what is the flow rate if the back pressure is 450 kPa? What are the velocities at the duct inlet and exit? Problem 6.4 A constant-area duct is connected to a converging-diverging nozzle (resembling those depicted in Figs. 6.6 and 6.7), which is in turn connected to a reservoir of air at 400 kPa and 400 K. The area ratio of the nozzle is 2.0 while the length to diameter ratio of the duct is 150 and the friction coefficient is 0.02. Find the maximum back pressure for the nozzle to be choked. Under this condition, what is the loss in stagnation pressure? Problem 6.5 For the same conditions given in Problem 6.4, except that the duct length to diameter ratio is 25, find the back pressure below which a normal shock exists and that the exit Mach number is 1 (downstream of the duct inlet). Also find the location of the shock. Problem 6.6 For the conditions given in Problem 6.4 (air reservoir pressure at 400 kPa, temperature at 400 K; nozzle area ratio of 2, duct length to diameter ratio of 150), find the back pressure for a normal shock to exist in the diverging section of the nozzle where the area ratio is1.30. What is the exit velocity? Problem 6.7 Following Example 6.2, if the stagnation pressure and stagnation temperature entering the duct are 500 kPa and 800 K, respectively, what are the exit pressure and velocity and what is the pressure loss?

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176

6 One-Dimensional Flows with Friction

Problem 6.8 Following the same conditions given in Example 6.7, except with a vacuum back pressure. Determine the location of the normal shock. Problem 6.9 Consider an air flow in an adiabatic constant-area rectangular duct (dimensions: 2 cm x 4 cm) connected to a reservoir (pressure = 300 kPa) through a converging-diverging nozzle (with an exit-to-throat area ratio of 1.688). The duct length is 50 cm and the friction factor f is 0.01. (a) Show that supersonic exit is possible. (b) What is the back pressure if no normal shock is to stand at the duct exit? (c) What is the highest back pressure for the nozzle-duct assembly to be choked? (d) What happens if the back pressure falls within the range between those determined in (b) and (c)?

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7

One-Dimensional Flows with Heat Transfer

Heat transfer in a compressible flow occurs because of the temperature difference between the flow and the surroundings across the non-adiabatic channel walls. Heat addition to the fluid can also arise from chemical, heat-releasing reactions that convert chemical enthalpy to thermal enthalpy. The flow in the convergingdiverging nozzle of a rocket is a good example, as the high-temperature flow may lose energy to the wall of the nozzle, while some chemical reactions may continue into the nozzle as they are not completed in the combustion chamber. Steady compressible one-dimensional flow in a frictionless constant-area channel is considered to focus on the significance of heat transfer. Similar to the Fanno line flow, the constant-area channel flow with gas supplied by a nozzle (converging or convergingdiverging) connected to a reservoir will be discussed. As in the Fanno line flow, a perfect gas is assumed.

7.1 Rayleigh Line Flow δq p

p + dp

ρ

ρ + dρ

T V M

T + dT V + dV M + dM

Figure 7.1 Steady one-dimensional flow in the channel of a constant cross-section area with frictionless walls and heat addition (i.e., a Rayleigh-line flow).

dx

The steady, inviscid one-dimensional compressible flow in the channel of a constant cross-section area, with heat transfer to the fluid and no external work, constitutes the Rayleigh line flow. To derive governing equations of the Rayleigh line flow consider the differential control volume shown in Fig. 7.1. The conservation of mass requires that

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178

7 One-Dimensional Flows with Heat Transfer

ρV ¼ ðρ þ dρÞðV þ dV Þ ¼ constant

(7.1)

By differentiating both sides of ρV ¼ constant and the dividing through by ρV, or by neglecting the second-order term in the equality ρV ¼ ðρ þ dρÞðV þ dV Þ, it is easy to show that dρ dV þ ¼0 ρ V

(7.2)

Because of the frictionless wall, the momentum conservation equation takes the form of Eqn. (2.24) p þ ρV 2 ¼ constant:

(2.24)

Noting that ρV ¼ constant, the momentum conservation equation simplifies to dp þ ρVdV ¼ 0

(7.3)

The energy conservation law with cp ¼ constant is

1 2 dht ¼ dðcp Tt Þ ¼ cp dTt ¼ dh þ d V ¼ cp dT þ VdV ¼ δq 2

(7.4)

where δq is the energy added to the fluid per unit mass. Equations (7.2), (7.3), and (7.4) are the governing equations for the Rayleigh line flow. Just as for the Fanno line flow, it is desirable to express these equations in terms of a Mach number. With p ¼ ρRT and a2 ¼ γRT, Eqn. (2.24) becomes   (7.5) p 1 þ γM2 ¼ constant pffiffiffiffiffiffiffiffiffi Because V ¼ M γRT , and p ¼ ρRT pffiffiffiffiffiffiffiffiffi p p pM γRT T¼ ¼ ¼ _ _ ρR ðm=AV ðm=A ÞR ÞR _ Therefore, with m=A ¼ ρV ¼ constant, pffiffiffiffi T ¼ ðconstantÞpM By using Eqn. (7.5) for p, this expression for T becomes T ¼ ðconstantÞ

M2 2

ð1 þ γM2 Þ

(7.6)

Equations (7.4), (7.5), and (7.6) provide the relations for p, T, and Tt between any two points in the Raleigh line flow: p2 1 þ γM12 ¼ p1 1 þ γM22

(7.7)

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7.1 Rayleigh Line Flow

179

 2 T2 M22 1 þ γM12 ¼   T1 M2 1 þ γM2 2 2 1

(7.8)

q ¼ cp ðTt2  Tt1 Þ

(7.9)

As in the Fanno line flow, it is convenient to adopt the properties at M ¼ 1 as the reference values. Thus, with location 1 designated as the reference state (M1 ¼ 1) and location 2 as an arbitrary location, the above equations can be rewritten as p 1þγ ¼ p 1 þ γM2

(7.10)

T ð1 þ γÞ2 M2 ¼ T  ð1 þ γM2 Þ2

(7.11)

The continuity requirement yields V ρ p =RT  ð1 þ γÞM2 ¼ ¼ ¼  V ρ p=RT 1 þ γM2

(7.12)

Locally isentropic relations

γ  1 2 γ=ðγ1Þ M pt ¼ p 1 þ 2 and

γ1 2 M Tt ¼ T 1 þ 2 lead to pt ¼ pt



1þγ 1 þ γM2





 2 γ  1 2 γ=ðγ1Þ M 1þ γþ1 2

(7.13)

and

Tt 2ð1 þ γÞM2 γ1 2 M ¼ 1þ 2 Tt ð1 þ γM2 Þ2

(7.14)

Numerical values of these ratios expressed in Eqns. (7.10) to (7.14) as functions of a Mach number are tabulated in Appendix G. The results are plotted in Fig. 7.2 for γ ¼ 1:4. For a better understanding of these relations, it is desirable to construct the T-s diagram for a given mass flow rate, much like the one for the Fanno line flow in Chapter 6.

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180

7 One-Dimensional Flows with Heat Transfer

104 103

pt /pt*

pt /pt*

102

ρ*/ρ 101 ρ*/ρ

Tt /Tt* 100

Tt /Tt*

T/T* −1 p/p* 10

T/ T* p/p*

10−2 10−3 10−4 0.01

0.1

1

10

M

Figure 7.2 Properties of Rayleigh line flow as a function of M for γ ¼ 1:4.

For an ideal gas, ds ¼ cp

dT dp R T p

Since R=cp ¼ ðγ  1Þ=γ s  s1 T ð γ  1Þ p ln ¼ ln  T1 γ p1 cp

(7.15a)

s  s T ðγ  1Þ p ln  ¼ ln   T γ p cp

(7.15b)

or, alternatively,

To obtain the T-s relation, the pressure term can be expressed in terms of temperature. Combining Eqns. (7.7) and (7.8) yields rffiffiffiffiffiffi p M1 T ¼ (7.15c) p1 M T1 Utilizing Eqns. (7.8) and (7.9) again produces    1  2 2 p1 1 þ γM1 M ¼ 1 γ p Substituting M from Eqn. (7.15d) into Eqn. (7.15c) yields sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi    p1 p 1  T 2 1 þ γM1  1 ¼ M1 p1 γ T1 p

(7.15d)

(7.15e)

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7.1 Rayleigh Line Flow

181

Squaring this expression and rearranging the result leads to 2

  p p T  1 þ γM12 ¼0 þ γM12 p1 p1 T1 Solving this quadratic equation results in sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi " #   2 p 1  T 2 2 2 1 þ γM1  ¼ 1 þ γM1  4γM1 p1 2 T1

(7.15f)

(7.15g)

Thus Eqn. (7.15b) for entropy change becomes sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( " ffi#)   2 s  s1 T ð γ  1Þ 1  T 2 2 2 ln 1 þ γM1  ¼ ln   1 þ γM1  4γM1 (7.16) T γ 2 T cp It is noted that the following inequality has to be satisfied for Eqn. (7.16):  2 1 þ γM12 T ≤ T1 4γM12 If the reference condition 1 is replaced by the sonic condition, then sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( " ffi#) s  s T ð γ  1Þ 1 T 2 ln ð1 þ γÞ  ð1 þ γÞ  4γ  ¼ ln   T γ 2 T cp

(7.17)

T T*

• • • m 3 < m2 < m1

B • m Cooling 1 Decelerating g

1

M
1 Δ−Δ∗ CP

Figure 7.3 Each T-s curve is the Rayleigh line for a given mass flow rate. The solid-line Rayleigh line curve (with mass flow rate m_ 1 ) demonstrates the effects of subsonic and supersonic heating and cooling on temperature and Mach number. The dashed-line curves show that for a given inlet condition, the effect of subsonic heating rates exceeding that required for Me ¼ 1(point A on the solid curve), resulting in reduced mass flow rates: m_ 3 < m_ 2 < m_ 1 . Downloaded from https:/www.cambridge.org/core. Columbia University Libraries, on 24 Jun 2017 at 04:19:32, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/9781316014288.008

182

7 One-Dimensional Flows with Heat Transfer

The T-s relationship described by Eqn. (7.17) is qualitatively sketched in Fig. 7.3, where the + and the – signs in Eqn. (7.16) are used respectively for the upper and the lower branches, separated by point B, where T is the maximum (T ¼ Tmax ) along the T-s curve. Similar to the Fanno line flow, there is a point for maximum entropy that also corresponds to M ¼ 1 conditions, as will be shown in the following, and is designated as Psonic . At Psonic , ds=dT ¼ 0. The continuity and momentum conservation throughout the flow require dρ dV þ ¼0 ρ V

(7.2)

dp þ ρVdV ¼ 0

(2.24)

Combining these two equations yields V2 ¼

dp dρ

At point Psonic , there is no variation in entropy. Thus

dp 2 V ¼ ¼ a2 dρ s That is, Mpsonic ¼ 1 at the point of maximum entropy, with the point already denoted as Psonic for easy identification. The condition for PTmax is dT ¼ 0. Thus, by using Eqn. (7.11), " # 2 2 h  2 1  ð1 þ γÞ M 2ð1 þ γÞ2 M 1 þ γM2 dT ¼ d T ¼ 2 2 2 2 ð1 þ γM Þ ð1 þ γM Þ  i ð1 þ γÞ2 M2 4γM 1 þ γM2 dM ¼ 0 Therefore, MTmax

sffiffiffi 1 ¼ γ

(7.18)

and

T T

¼ max

ð1 þ γÞ2 4γ

It is apparent that MTmax < 1, and for air (γ ¼ 1:4), ðT=T  Þmax ¼ 0:735.

(7.19) MTmax ¼ 0:845 and

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7.2 Reference State in Rayleigh Line Flow

183

To understand more clearly the general characteristics of the T-s curve, it is instructive to express the entropic change as a function of a Mach number. Substituting Eqns. (7.8) and (7.9) into Eqn. (7.15a) yields "  2 #   M2 1 þ γM12 1 þ γM12 s  s1 T ðγ  1Þ p ð γ  1Þ ln ¼ ln ln ¼ ln   2 T1 γ p1 γ cp 1 þ γM2 M12 ð1 þ γM2 Þ "

ðγþ1Þ=γ # M 2 1 þ γM12 (7.20) ¼ ln M1 1 þ γM2 Because the maximum temperature occurs under the subsonic condition, the upper branch of T-s curve shown in Fig. 7.3 is the subsonic branch, while the lower branch is the supersonic branch. On the subsonic branch, heat addition increases the entropy and accelerates the flow toward the sonic condition (point Psonic ); on the supersonic branch heat addition increases entropy and leads to deceleration toward Psonic . There are alternative, qualitative ways of ascertaining the flow acceleration/deceleration resulting from heat transfer; Problem 7.11 provides the instruction and exercise for obtaining the results.

7.2 Reference State in Rayleigh Line Flow If M1 ¼ 1 is the reference state, then Eqn. (7.20) becomes "

ðγþ1Þ=γ # s  s 1 þ γ ¼ ln M2 1 þ γM2 cp  It can be seen that because ðγ þ 1Þ=γ > 1, M

2




1þγ 1þγM2

ðγþ1Þ=γ 

(7.21) 

and therefore ss cp are

increasing (decreasing) functions of M for M < 1 ðM > 1Þ. It can be concluded that, for a given flow rate (i.e., a given T-s curve) with heat addition, an initially subsonic flow accelerates, while an initially supersonic flow decelerates toward the sonic condition (point Psonic on the T-s curve). With heat removal from the flow, the opposite occurs. Because

T T

2

1 ¼ ð1þγÞ M , it is also concluded that the subsonic 2 2 e M2 2

ð1þγM Þ

branch of the T-s curve has larger values of TT than the supersonic branch; so the subsonic regime corresponds to the upper branch. This conclusion is consistent with the result shown in Eqn. (7.18) in that PTmax is on the upper branch and pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi MTmax ¼ ð1=γÞ < 1. As can be seen in Fig. 7.3, for M < ð1=γÞ on the subsonic branch of the T-s curve, T=T  increases with increasing M by heat addition (or pffiffiffiffiffiffiffiffiffiffiffi simply, heating); for 1 > M > ð1=γÞ, T=T  decreases with increasing M still by heating. For M > 1(the lower, supersonic branch), T=T  always increases as M is decreased by heating. (A quantitative description of the trends of the combined effects of Mach number and heat transfer can be derived and is left as an end-ofchapter exercise, Problem 7.11.)

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184

7 One-Dimensional Flows with Heat Transfer

As with the Fanno line flow, similar questions are asked for the Rayleigh line flow as to what occurs when the amount of heat addition exceeds that required to accelerate/decelerate a subsonic/supersonic flow to the sonic condition. For subsonic Rayleigh line flow (which can be generated by connecting the constant-area duct to a reservoir by a converging nozzle), the inlet flow to the channel is subsonic because of the converging nozzle. By heating, the flow in the constant-area channel cannot become supersonic, according to the result shown in Fig. 7.3. If heat addition exceeding that required for M ¼ 1, the mass flow rate will be reduced as the flow is affected by the downstream heating and a new Rayleigh line is established, as shown in Fig. 7.3. In Fig. 7.3, m_ 2 < m_ 1 , where m_ 1 is for subsonic flow with the heat addition and m_ 2 is the mass flow rate due to excessive heat addition. Even more heat addition results in yet another Rayleigh line with m_ 3 < m_ 2 . Therefore, a duct flow can be choked by heat addition, and the maximum amount of heat that can be added to the flow for a given _ is determined by the attainment of the sonic condition. Rayleigh line (i.e., a given m) Similar to the Fanno line flow, whether the choking condition occurs also depends on the back pressure. For example, if the back pressure is sufficiently low, then the exit Mach number Me ¼ 1, and the stagnation and static temperatures attained at the exit are Tt and T  , respectively. This is illustrated by curves 4–6 in i

δq

e

Pr Mi < 1

Tr

Pb

P

P =1 Pr

(1)

Pe = Pb (2) Mi = 1

Me < 1

(3)

(4) (5) (6)

Pe > Pb Me = 1 x

Figure 7.4 Rayleigh-line flow in a converging nozzle followed by a constant-area duct. For pb ≤ pð4Þ the sonic exit condition is established at the duct exit and the flow is thermally choked; _ does not increase for pb ≤ pð4Þ ; for pb ≥ pð4Þ , m_ increases as pb that is, the mass flow rate, m, decreases. For pb < pð4Þ , the flow continues to expand outside the duct.

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7.2 Reference State in Rayleigh Line Flow

185

Fig. 7.4. Therefore, the flow is subsonic throughout the nozzle-channel system, with the possibility of achieving sonic exit conditions if pb is sufficient. Curve 4 corresponds to the maximum back pressure that would choke the flow (with Me ¼ 1 and pe ¼ pb ¼ p ) for the given mass flow rate and heat addition. Above this back pressure, the exit condition is subsonic, represented by curves 1–3. For back pressure lower than that of curve 4 (i.e., curves 5 and 6), Me ¼ 1, pe > pb , and the flow continues to expand outside the duct exit. For a supersonic Rayleigh flow, the adjustment to excessive heating results in a normal shock wave in the duct. The location of the normal shock wave depends on the back pressure and the distribution of the head addition along the length of the duct. Because the amount of heat transfer varies with the flow speed and the gas temperature, unlike the friction coefficient that is constant throughout the Fanno line flow, the determination of the shock location is more complicated than that in the Fanno line flow. This scenario is schematically shown in Fig. 7.5. δq

Mth = 1

Pr Tr

1 Mi > 1

2

M1 > 1 M2 < 1

Pe Me < 1

Normal shock P =1 Pr Pe = Pb

x

Figure 7.5 Rayleigh-line flow in a converging-diverging nozzle followed by a constant-area duct with duct inlet Mach number Mi > 1 and δq > 0 such that a normal shock stands within the duct. Note that Me < 1 and pe ¼ pb .

There exists a significant difference between the Fanno and Rayleigh line flows regarding entropy changes. In a Fanno line flow, entropy can only increase in the direction of the flow. However, entropy can either decrease or increase in a Rayleigh line flow, depending on whether heat addition or heat loss takes place in the fluid. One distinct example is that if heat loss occurs in the converging nozzle-duct assembly, a favorable back pressure would result in a sonic condition at the entrance flow to the duct, which would accelerate to become a supersonic flow throughout the duct. For

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186

7 One-Dimensional Flows with Heat Transfer

such an assembly of a converging nozzle and a constant-area duct, the Fanno line flow cannot reach supersonic conditions, as already discussed in Chapter 6. The net result of heat transfer is best seen in the change in stagnation temperature, as indicated by Eqn. (7.9): q ¼ cp ðTt2  Tt1 Þ. The reference stagnation temperature Tt at M ¼ 1 can be calculated using Eqn. (7.14), based on conditions at a given location where M≠1 (i.e., Tt1 ). The new Mach number can then be found by   the ratio of Tt2 =Tt , followed by new flow parameters (at the new location 2) calculated using Eqns. (7.10) through (7.13). Air supplied from a reservoir through a converging nozzle enters a frictionless circular duct with a Mach number M1 of 0.5 and a temperature and a pressure of 300 K and 300 kPa, respectively. Over the length, heat was added to the flow at 100 kJ/kg. Find the mass flow rate and the exit velocity at the exit. What is the back pressure? What is the change in entropy? Assume the specific heat of air is 1.004 kJ/kg⋅K.

EXAMPLE 7.1

Solution – k p1 300 kPa For M1 ¼ 0:5, TTt11 ¼ 0:9542 ¼ 300 Tt1 ; pt1 ¼ 0:843 ¼ pt1 Therefore, Tt1 ¼ 314:4 K and pt1 ¼ 355:9 kPa. Also using the tabulated values of Appendix G, TTt1 ¼ 0:6914, ppt1 ¼ 1:1141. t t Therefore Tt ¼ 454:7 k and pt ¼ 319:5 kPa. q ¼ 100 kJ=kg ¼ cp ðTt2  Tt1 Þ ¼ 1:004 kJ=kg ⋅ K  ðTt2  314:4 KÞ. Therefore, Tt2 ¼ 414:0 K. p2 pt2 Tt1 T2 T  ¼ 0:9105. So, M2 ≈ 0:70, for which Tt2 ¼ 0:9107, pt2 ¼ 0:7209, and p ¼ 1:0431 t

t

pt2 p 1  355:9 kPa ¼ 333:2 kPa pt2 ¼   t  pt1 ¼ 1:0431  1:1141 pt pt1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi T2 ¼ 373:3 K and V2 ¼ M2 γRT2 ¼ 0:70  1:4  287  373:3 ¼ 271:1 m=s. p2 ¼ ppt22  pt2 ¼ 0:7209  333:2 kPa ¼ 240:2 kPa. Therefore, p2 ¼ pb ¼ 240:2 kPa The mass flow rate is p2 pffiffiffiffiffiffiffiffiffiffiffi 240:2  103 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kg m_  1:4  287  373:3 ¼ 605:0 2 ¼ ρ2 V2 ¼ γRT2 ¼ m ⋅s RT2 287  373:3 A Alternatively, M1 ¼ 0:5 leads to pp1 ¼ 1:7778. M2 ≈ 0:70 gives pp2 ¼ 1:4235  1  300 ¼ 240:2 kPa Then, p2 ¼ pb ¼ pp2 pp1 p1 ¼ 1:4235  1:7778 For steady-state flow, mass flow rate can be calculated at any location. For now at the duct entrance, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 300  103 p1 m_  0:5  1:4  287  300 ¼ ρ1 V1 ¼ M1 γRT1 ¼ RT1 287  300 A ¼ 604:9 kg=s ⋅ m2 The above conditions correspond to the conditions represented by either curve 1 or curve 2 in Fig. 7.3.

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7.2 Reference State in Rayleigh Line Flow

s2  s1 ¼ ln cp

187

" " #

ðγþ1Þ=γ # M2 2 1 þ γM12 0:7 2 1 þ 1:4  0:25 2:4=1:4 ¼ ln 0:5 1 þ 1:4  0:49 M1 1 þ γM22 ¼ 0:292

□

Use the nozzle-duct assembly from Example 7.1 with reservoir pressure and temperature equal to 355:9 kPa and 314:4 K, respectively, and q ¼ 100 kJ=kg. What is the mass flow rate and what is the exit velocity if Pb ¼ 270 kPa?

EXAMPLE 7.2

Solution – Unlike Example 7.1, where M1 is known to allow easy determination of the reference conditions, the current problem is solved by iterations. First guess: M1 ¼ 0:42. Therefore tabulated values of Appendix G indicate TTt1 ¼ 0:5638 and t pt1 pt ¼ 1:148. Therefore Tt ¼ 557:6 k and pt ¼ 310:0 kPa. q ¼ 100 kJ=kg ¼ cp ðTt2  Tt1 Þ ¼ 1:004 kJ=kg ⋅ K  ðTt2  314:4 KÞ. Therefore, Tt2 ¼ 414:0 K. Tt2 T  ¼ 0:7425 So, M2 ≈ 0:54, for which t

T2 p2 pt2 p2 ¼ 0:9449; ¼ 0:8201;  ¼ 1:0979; and  ¼ 1:7043 Tt2 pt2 pt p pb ¼ p2 ¼

p2 pt2 pt 1  355:9kPa    pt1 ¼ 0:8201  1:0979  1:148 pt2 pt pt1 ¼ 279:1 kPa

Second guess: M1 ¼ 0:44. Therefore, TTt1 ¼ 0:5979 and ppt1 ¼ 1:1394. t t Therefore, Tt ¼ 525:8 k and pt ¼ 312:4 kPa. q ¼ 100 kJ=kg ¼ cp ðTt2  Tt1 Þ ¼ 1:004 kJ=kg ⋅ K  ðTt2  314:4 KÞ. Therefore, Tt2 ¼ 414:0 K. p2 pt2 Tt2 T2 Tt ¼ 0:7874. So, M2 ≈ 0:57, for which Tt2 ¼ 0:940, pt2 ¼ 0:80, pt ¼ 1:086, and p2 p ¼ 1:65 p2 ¼

p2 pt2 pt 1  355:9 kPa ¼ 271:5 kPa    pt1 ¼ 0:80  1:086  1:139 pt2 pt pt1

For answers to the questions, M1 ¼ 0:44 and M2 ≈ 0:57. p1 T1 pt1 ≈ 0:8755 and Tt1 ¼ 0:9627; thus, p1 ¼ 311:6 kPa and T1 ¼ 302:7 K pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 311:6  103 p1 m_ ¼ ρ1 V1 ¼  0:44  1:4  287  302:7 M1 γRT1 ¼ A RT1 287  302:7 ¼ 550:4 kg=s ⋅ m2

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188

7 One-Dimensional Flows with Heat Transfer

V 2 ¼ M2

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi γRT2 ¼ 0:57  1:4  287  0:94  414 ¼ 225:4 m=s

Comments – When the duct is not choked, an increase in the back pressure leads to decreases in both the mass flow rate and exit velocity. Therefore, the results in Examples 7.1 and 7.2 can be qualitatively represented by curves 2 and 1, □ respectively, in Fig. 7.4. Repeat Example 7.1 except that, instead of heat addition, now heat is taken away from the flow at 100 kJ/kg.

EXAMPLE 7.3

Solution – K p1 300 kPa For M1 ¼ 0:5, TTt11 ¼ 0:9542 ¼ 300 Tt1 ; pt1 ¼ 0:843 ¼ pt1 Therefore, Tt1 ¼ 314:4 K and pt1 ¼ 355:9 kPa Also, TTt1 ¼ 0:6914, ppt1 ¼ 1:1141. t t Therefore, Tt ¼ 454:7 K and pt ¼ 319:5 kPa. q ¼ 100 kJ=kg ¼ cp ðTt2 Tt1 Þ ¼ 1:004 kJ=kg ⋅ K  ðTt2  314:4 KÞ. Therefore, Tt2 ¼ 214:8 K. p2 pt2 Tt2 T2 T  ¼ 0:4724 So, M2 ≈ 0:37, for which Tt2 ¼ 0:973, pt2 ¼ 0:910, and p ¼ 1:169 t

t

p2 ¼ 0:91 pt2 p

1 pt2 ¼ ppt2  pt1t  pt1 ¼ 1:169  1:1141  355:9 kPa ¼ 373:4 kPa t

p2 ¼ 0:91  373:4 kPa ¼ 339:8 kPa pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi T2 ¼ 209:0 K and V2 ¼ M2 γRT2 ¼ 0:37  1:4  287  209:0 ¼ 107:2 m=s. p2 pffiffiffiffiffiffiffiffiffiffiffi 339:8  103 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kg m_  1:4  287  209:0 ¼ 605:8 2 γRT2 ¼ ¼ ρ2 V2 ¼ m ⋅s RT2 287  209:0 A " " #



ðγþ1Þ=γ # s2  s1 M2 2 1 þ γM12 0:37 2 1 þ 1:4  0:25 2:4=1:4 ¼ ln ¼ ln 0:5 1 þ 1:4  0:1369 cp M1 1 þ γM22 ¼ 0:388 Comments – Cooling results in an increase in the stagnation pressure and a decrease in entropy at the exit, contrary to heating as shown in Example 7.1. However, the mass flow rate remains the same as it can be determined by the conditions at the duct inlet, which are the same, but can now exit to a higher back pressure. These □ results are consistent with the T-s curve in Fig. 7.2. To choke the nozzle-duct assembly in Example 7.1 with the same reservoir (not inlet) conditions, what is the amount of heat addition to cause the flow to be sonic? What is the maximum back pressure if the heat addition is (1) 300 kJ/kg and (2) 500 kJ/kg?

EXAMPLE 7.4

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7.2 Reference State in Rayleigh Line Flow

189

Solution – Again, The reservoir conditions are Tr ¼ Tt1 ¼ 314:4 K and pr ¼ pt1 ¼ 355:9 kPa To reach sonic conditions at the duct exit, the total amount of heat addition is   kJ kJ  ð454:7 K  314:4 KÞ ¼ 140:9 q ¼ cp Tt  Tt1 ¼ 1:004 kg ⋅ K kg By adding a larger amount of heat that this value results in the duct being choked at the exit and Me ¼ M2 ¼ 1. (1) For q ¼ 300 kJ=kg > 140:9 kJ=kg, so Me ¼ 1, but M1 will no longer be 0.5 and there will be different values of Tt and pt from that in Example 7.1. Tt ¼ Tt1 þ

q 300 kJ=kg ¼ 613:2 K ¼ 314:4 K þ cp 1:004 kJ=kg ⋅ K

Therefore, TTt1 ¼ 314:4 613:2 ¼ 0:5127 and the new Mach number is M1 ≈ 0:39, for t which ppt11 ≈ 0:90, ppt1 ¼ 1:16, and pt ¼ 306:8 kPa. t

T1 Tt1

≈ 0:97. Therefore, p1 ¼ 0:90  355:9 kPa ¼ 319:5 kPa and T1 ¼ 305:0 K

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 319:5  103 p1 m_  0:39  1:4  287  305:0 ¼ ρ1 V1 ¼ M1 γRT1 ¼ RT1 287  305:0 A ¼ 498:3 kg=s ⋅ m2 : 

Now because Me ¼ 1, pptee ¼ pp ¼ 0:5283. For M1 ≈ 0:39, ppt1 ≈ 1:16. t t  p 1 Therefore, pe ¼ pp  pt1t  pt1 ¼ 0:5283  1:16  355:9 kPa ¼ 162:1 kPa t _ 2 For m A ¼ 498:3 kg=s ⋅ m and q ¼ 300 kJ=kg , pb < 161:9 kPa will ensure a choked duct (corresponding to curve 3 in Fig. 7.3). Alternatively, pp1 ¼ 1:979.  1 pe ¼ p ¼ pp1  p1 ¼ 1:979  319:5 kPa ¼ 161:4 kPa ≈ 161:9 kPa (as expected – the two approaches for the static should yield the same result). 500 kJ=kg (2) For q ¼ 500 kJ=kg, Tt ¼ Tt1 þ cqp ¼ 314:4 K þ 1:004 kJ=kg ⋅ K ¼ 812:4 K Tt1 314:4 Therefore, T  ¼ 812:4 ¼ 0:387 and the new Mach number is M1 ≈ 0:32, for which t p1 T1 pt1 ≈ 0:9315 and Tt1 ≈ 0:980. Therefore, p1 ¼ 0:9315  355:9 kPa ¼ 331:5 kPa and T1 ¼ 308:1 K pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 331:5  103 p1 m_  0:32  1:4  287  308:1 ¼ ρ1 V1 ¼ M1 γRT1 ¼ RT1 287  308:1 A ¼ 422:1 kg=s ⋅ m2 : 

Now because Me ¼ 1, pptee ¼ pp ¼ 0:5283. For M1 ≈ 0:32, ppt1 ≈ 1:190. t t  p 1  355:9 kPa ¼ 158:0 kPa Therefore, pe ¼ pp  pt1t  pt1 ¼ 0:5283  1:190 t _ 2 and q ¼ 500 kJ=kg, pb < 158:0 kPa will ensure For m A ¼ 422:1 kg=s ⋅ m a choked duct (corresponding to curve 3 in Fig. 7.3). Alternatively, pp1 ¼ 2:0991 and  1  331:5 kPa ≈ 158:0 kPa pe ¼ p ¼ pp1  p1 ¼ 2:0991

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190

7 One-Dimensional Flows with Heat Transfer

Comments – 1. This example illustrates that increasing the amount of heat addition, the choked condition results in a reduction in the mass flow rate and the point Psonic moves to the right, as depicted in Fig. 7.3. 2. In Examples 7.1 and 7.2, specifying M1 for the given reservoir pressure and temperature (albeit calculated based on the condition at the entrance of the duct) indirectly specifies pe , which is equal to pb , and vice versa. This is because the amount of heat addition is not sufficient to accelerate the flow to sonic conditions at the duct exit and the exit pressure has to equilibrate to that in the surrounding. In the current example, there exists a corresponding back pressure for a given _ do amount of heat addition. Below this critical back pressure, both M1 and m=A □ not change further, because the duct is choked. Air supplied from a reservoir (the reservoir temperature and pressure are 400 K and 400 kPa, respectively) through a converging nozzle enters a frictionless circular duct, as depicted in Fig. 7.3. Over the length, heat was added to the flow at 500 kJ/kg. Find the mass flow rate and the exit velocity if the back pressure is 275 kPa. Assume the specific heat of air is 1.004 kJ/kg⋅K.

EXAMPLE 7.5

Solution – First of all, it is important to decide whether the duct is thermally choked by heat transfer. If it is, then Me ¼ 1 and pe ¼ p ¼ pb ¼ 275 kPa is the maximum back pressure for choking (curve 4 in Fig. 7.3). Therefore, pe pe   pte ¼ 0:5238 ¼ pt . Thus pt ¼ 515 kPa and pe ¼ p . Since during subsonic acceleration pt can only decrease, it cannot exceed the reservoir. Therefore, the duct must not be choked and Me < 1(the condition resembles that represented by curve 1 or 2 in Fig. 7.3). One way to solve the problem is to guess the duct inlet Mach number and then, based on the change in Tt , to find Me and pe . If pe ¼ pb , the solution is found. Another path to a solution is to assume that the exit condition is sonic. There are two approaches. The first is described as follows: First guess – the inlet Mach number Mi ¼ 0:3, then ppti ¼ 1:1985 and t Tti  Tt ¼ 0:3469 and therefore Tt ¼ 400 K=0:3469 ¼ 1; 153:1 K. Tte ¼ Tti þ q=cp 898 ¼ 400 þ 500=1:004 ¼ 898:0 K. TTte ¼ 1;153 ¼ 0:7788. t pe Therefore, Me ≈ 0:56, pte ¼ 0:8082 and ppte ¼ 1:0901. t p 1  400 kPa ¼ Therefore, pe ¼ pptee  ppte  ptit  pti ¼ 0:8082  1:0901  1:1985 t 294 k Pa > pb Second guess – the inlet Mach number Mi ¼ 0:34, then ppti ¼ 1:1822 and t Tti  Tt ¼ 0:4206 and therefore Tt ¼ 400 K=0:4206 ¼ 951:0 K. Tte ¼ Tti þ q=cp ¼ 898 ¼ 0:9443. 400 þ 500=1:004 ¼ 898:0 K. TTte ¼ 951:0 t pe pte Therefore, Me ≈ 0:63, pte ¼ 0:765 and p ¼ 1:065. t p 1  400 kPa ¼ Therefore, pe ¼ pptee  ppte  ptit  pti ¼ 0:765  1:065  1:1822 t 275:6 k Pa ≈ pb .

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7.2 Reference State in Rayleigh Line Flow

Thus Mi ¼ 0:34, Me ≈ 0:63, pi pe p ¼ 2:0657, and p ¼ 1:543.

pi pti

¼ 0:9231,

pe pte

¼ 0:765,

Ti Tti

¼ 0:9774,

191 Te Tte

¼ 0:923,

pffiffiffiffiffiffiffiffiffiffi 0:9321  400  103 pi m_  0:32 ¼ ρi Vi ¼ Mi γRTi ¼ RTi 287  0:9774  400 A pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  1:4  287  0:9774  400 ¼ 447:8 kg=s ⋅ m2 pffiffiffiffiffiffiffiffiffiffiffi Ve ¼ Me γRTe ¼ Me

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Te Te ¼ 0:34  1:4  287  0:923  898:0 γR Tte ¼ 196:2 m=s:

The second approach to a solution is to assume that the exit condition is sonic with pe ¼ p ¼ pb ¼ 275 kPa, and pt ¼ pte . Under this assumption pt ¼ pte ¼ pe =0:5283 ¼ 525:0kPa > pr , which is not possible. Therefore, Me < 1 and the □ previous solution must be followed. A model supersonic combustor has the air flow at the inlet at Mach number 2.5 and temperature and pressure at 226 K and 21 kPa, respectively. Assume a Rayleigh line flow in the combustor and the average specific heat is cp ¼ 1:004 ⋅ kJ=kg ⋅ K. What amount of heat addition is needed to have the Mach number to be unity at the combustor exit? Under this condition, what is the loss in stagnation pressure across the combustor? If the JP8 fuel adds 1,900 kJ/kg to the gases, what may happen in the combustor? EXAMPLE 7.6

Solution – For Mi ¼ 2:5, TTtii ¼ 0:4444, pptii ¼ 0:0585, ppti ¼ 2:2218, and TTti ¼ 0:7101 t t Therefore, Tti ¼ Ti =0:444 ¼ 226=0:444 ¼ 508:6 K, Tt ¼ Tti =0:7101 ¼ 716:2 K. cp ðTte  Tti Þ ¼ 1:004  ð716:2 K  508:6 KÞ ¼ 208:4 kJ=kg: Under the choked condition,



pt pti 1 1 1   21 kPa pte  pti ¼ pt  pti ¼ 1  pi ¼ 2:2218 0:0585 pti pi ¼ 197:4 kPa For q ¼ 1; 900 kJ=kg, Tte ¼ 2401:0 K, then TTte ¼ 2;401:0 716:2 > 1. So the heat addit tion is more than needed to decelerate the flow to sonic conditions. Therefore, □ there must be a normal shock in the combustor. A constant area duct is connected to an air reservoir (at 700 kPa and 400 K) through a converging-diverging nozzle. The area ratio of the duct inlet to the throat is 2.0 and the heat addition throughout the duct is 50 kJ/kg. Determine the range of back pressure for a normal shock wave to stand within the duct.

EXAMPLE 7.7

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192

7 One-Dimensional Flows with Heat Transfer

Solution – Assume that the nozzle is isentropic. Therefore, a 2.0 area ratio leads to Mi ¼ 2:20, Pti ¼ Pr ¼ 700 kPa, Tti ¼ Tr ¼ 400 k, and Ppti ¼ 1:7434, and t Tti Tt ¼ 0:7561. (Let subscripts 1 and 2 denote locations immediately upstream and downstream of the shock, respectively.) First scenario – The normal shock stands at the exit plane T þq=c Pt1 Therefore, TTt1 ¼ Tti =T Tp ¼ 400þ50=1:004 400=0:7561 ¼ 0:8502 and so M1 ¼ 1:74, pt ¼ 1:2692. t ð t ti Þ ti Across the shock wave, M2 ¼ 0:6305, PPt2t1 ¼ 0:8389. Also TTt22 ≈ 0:9265 and ppt22 ≈ 0:7653 p2 ¼ pb ¼

p2 Pt2 Pt1 pt 1     Pti ¼ 0:7653  0:8389  1:2692  1:7434 pt2 Pt1 pt Pti 700 kPa ¼ 327:2 kPa

Second scenario – The normal shock stands at the inlet of the duct Mi ¼ M1 ¼ 2:20, so M2 ¼ 0:5471 and TTt2 ¼ TTt1 ¼ TTr ¼ 0:756, Ppt2 ≈ 1:095 t t t t Tt2 þq=cp 400þ50=1:004 Tte Pte Tt ¼ ðT  =Tt2 ÞTt2 ¼ 400=0:756 ¼ 0:8502, and therefore, Me ≈ 0:627, pt ≈ 1:068 and t pe pte ¼ 0:927 T

Fanno M2 < 1 2

igh

yle Ra

1 M1 > 1

Δ

Figure 7.6 The intersection of the Rayleigh and Fanno lines determines the normal shock states. The dashed line signifies that the process from supersonic (point 1) to subsonic (point 2) is non-isentropic.

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7.3 Normal Shock on Rayleigh and Fanno Line T-s Diagrams

193

Across the shock PPt2t1 ¼ 0:6281. Therefore, pe ¼ pb ¼

pe Pte pt Pt2      Pt1 pte pt Pt2 Pt1

¼ 0:927  1:068 

1  0:6281  700 kPa ¼ 397:5 kPa 1:095

The range of back pressure for the normal shock to stand within the duct is 327:2 kPa ≤ pb ≤ 397:5 kPa (As the back pressure increases, the location of the normal shock moves □ upstream toward the nozzle.)

7.3 Normal Shock on Rayleigh and Fanno Line T-s Diagrams Recall that to develop the T-s diagram for Fanno line flow, the continuity and energy equations were used: ρV ¼ constant

(6.1)

1 ht ¼ cp Tt ¼ h þ V 2 ¼ constant 2

(6.2)

For the Rayleigh line flow, the continuity, Eqn. (6.1) and momentum equations were used: p þ ρV 2 ¼ constant

(7.24)

These three conservation equations were used to derive normal shock relationships. For a constant-area channel flow with heat transfer and wall friction, if a normal shock exists in the duct, then the states immediately upstream and downstream of the shock must appear at the intersection of the of the T-s curves of the Fanno and Rayleigh lines. This result for a given mass flow rate is schematically shown in Fig. 7.6. Since entropy increases across the shock, state 2 (downstream; M2 < 1) lies to the upper right of state 1 (upstream; M2 > 1) with T2 > T1 and s2 > s1 , due to the non-isentropic nature of shocks. The line connecting the two states is dashed because the shock process is not a thermodynamic equilibrium process and its path is not exactly known. Although the two T-s curves in Fig. 7.6 are for a given mass flow rate, it does not specify the exact combination of the heat addition and friction that leads to the normal shock conditions. For the conditions immediately upstream and downstream of the normal shock, the three governing equations have to be considered simultaneously. The following section is devoted to such a purpose.

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194

7 One-Dimensional Flows with Heat Transfer

7.4 Flow with Friction and Heat Addition The differential forms of conservation of mass and momentum for the constant-area duct are: dρ dV þ ¼0 ρ V

(6.16)

dp 1 dx dV þ γM2 f ¼0 þ γM2 p 2 Dh V

(6.14)

For energy conservation,




δq ¼ cp dTt

2 Since Tt ¼ T 1 þ γ1 2 M , the following expression is found:

δq γ  1 2 dT M þ ðγ  1ÞMdM ¼ 1þ cp T 2 T

(7.22)

For an ideal gas, dp dρ dT ¼ þ p ρ T

(6.15)

After substituting the continuity into this expression, dp dV dT ¼ þ p V T

(7.23a)

By using M ¼ pVffiffiffiffiffiffiffi, it is readily shown that γRT

dV dM 1 dT ¼ þ V M 2 T

(7.23b)

dp dM 1 dT ¼ þ p M 2 T

(7.23c)

Thus

Substitution of these two expressions into Eqn. (6.14) yields  dT  2  dM 1 1 dx 1 þ γM2 þ γM  1 þ γM2 f ¼0 2 T M 2 Dh

(7.23d)

After using dT=T from Eqn. (7.22) and some algebraic manipulations, one obtains "   #  δq 1  M2 1 1 dM 2 2 dx 1 þ γM (7.24) þ γM f ¼ ðγ1Þ 2 2 cp Tt 2 Dh 1þ M M 2

Therefore, with δq ¼ cp dTt

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7.5 Flow with Friction, Heat Transfer, and Area Change

" #  Þ 2    1 þ ðγ1 dM 1 1 2 1 dTt 2 2 M ¼M 1 þ γM þ γM f dx 2 Tt dx 2Dh ð1  M2 Þ

195

(7.25)

Once Tt (i.e., δq) as a function of x is known, Mach number at any location along the duct can be determined by integrating Eqn. (7.25).

7.5 Flow with Friction, Heat Transfer, and Area Change The flow in a rocket nozzle simultaneously experiences friction, heat transfer, and area change (Hill and Peterson, 1991; Sutton and Biblarz, 2001). The effects of friction and area change are apparent. Heat transfer may result from the remaining chemical reactions due to the shift in thermodynamic equilibrium as the nozzle flow creates changes in both temperature and pressure, to which chemical reaction and equilibrium conditions are sensitive. Heat transfer may take place simply due to the temperature difference between the gas and the nozzle wall, with the latter being exposed to the surroundings that may have vastly different temperatures. Assume that the flow is one-dimensional, and the continuity, momentum, and energy equations are, respectively, dρ dV dA þ þ ¼0 ρ V A

(5.2)

dp 1 dx dV þ γM2 f ¼0 þ γM2 p 2 Dh V

(6.14)

or dp þ ρVdV ¼ 0

(7.3)

δq ¼ cp T



γ  1 2 dT M þ ðγ  1ÞMdM 1þ 2 T

Because of the area change, Eqn. (6.15) is modified to be

dp dV dA dT ¼ þ þ p V A T

(7.22)

(7.26)

Substituting Eqn. (7.26) into Eqn. (6.14), or simply adding dA=A to Eqn. (7.23d), yields  dT  2  dM 1 1 dx dA 1 þ γM2 þ γM  1 þ γM2 f ¼0  2 T M 2 Dh A

(7.27)

Similarly,

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196

7 One-Dimensional Flows with Heat Transfer

"   #  δq 1  M2 1 1 dA dM 2 2 dx 1 þ γM ¼ þ γM f  Þ 2 2 cp Tt 2 Dh A M 1 þ ðγ1 M 2

(7.28a)

or " #  Þ 2   1 þ ðγ1 dM2 dA  2 dTt 2 dx 2 M þ 1 þ γM ¼ þ γM f 2 A Dh M2 ð1  M 2 Þ Tt

(7.28b)

Eqn. (7.28a) can be rewritten for Mach number variation along the flow direction, as " #  Þ 2   1 þ ðγ1 dM 1 dA 1  1 γM2 f 2 1 dTt 2 M ¼M þ 1 þ γM þ (7.29)  dx A dx 2 Tt dx 2 Dh ð1  M 2 Þ It is clear that from Eqn. (7.29) that the effects of area change, heat transfer, and friction all contribute to determining the location where sonic conditions exist. This is because for the right-hand side of Eqn. (7.29) to be meaningful for M ¼ 1, the following condition must be satisfied:    dA 1  2 dTt 2 dx ¼ 1 þ γM þ γM f (7.30) A 2 Dh M¼1 Tt That is, the location for M ¼ 1 is no longer at the throat (i.e., dA=dx ¼ 0) when nonisentropic effects such as friction and heat transfer are included. The exact location for M ¼ 1 depends on the relative magnitudes of the heat transfer and friction. Since the friction term is always positive, its contribution is to shift the sonic location to where dA=dx > 0, that is, the diverging section downstream of the throat. Depending on whether heat is lost/gained by the fluid, the effect of heat transfer is to shift the sonic location upstream/downstream of the throat. If the sum of contributions from friction and heat transfer is positive/negative, the sonic location is situated downstream/upstream of the throat. For a constant-area duct flow with heat transfer and friction, Eqn. (7.29) is reduced to Eqn. (7.25). The general solution of Eqn. (7.29) for M ¼ Mð xÞ may be obtained using numerical integration methods for given dA=dx, dTt =dx, and f as functions of x.

(a) a

Unburned (fresh) Vc gas, Vg = 0

b

Burned gases (Products) X

(b) a V1 = Ve

b V2 = V1–Vb

Figure 7.7 (a) A long tube with one closed end containing a reactive mixture initially at rest; the figure shows a combustion/pressure wave propagating into the quiescent mixture sometime after ignition is initiated at the closed end; (b) the flow field as observed by riding on the wave.

X

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7.6 Detonation and Deflagration Waves

197

7.6 Detonation and Deflagration Waves A special case of heat addition involves heat-releasing chemical reactions (Glassman and Yetter, 2008). Consider a long tube with one end closed that is filled with reactive mixture, where the fresh gas mixture is initially at rest. After the combustion reaction is initiated at the closed end, the energy release raises the temperature of the burned gas, igniting the adjacent layer of fresh mixture. A selfsustaining combustion wave is generated, as shown in Fig. 7.7a. At the same time, the thermal expansion of the burned gas (due to combustion) results in a pressure wave that pushes into the fresh mixture while carrying the hot product gas to the right, similar to a piston motion that may lead to a shock wave. For reactions with high activation energies, the combustion wave is thin, the temperature gradient is large across the combustion wave, and the pressure and combustion waves coincide. After an initial transient period, the pressure wave propagates with a steady-state speed, denoted by Vc . It is useful to know about the propagation speed of the combustion wave Vc . Assuming a steady-state one-dimensional wave propagation, Eqns. (4.66) through (4.69) for conservation of mass, momentum, and energy can be rewritten in the coordinate system following the motion of the wave (Fig. 7.7b) as ρ1 V1 ¼ ρ2 V2

(7.31)

ρ1 V12 þ p1 ¼ ρ2 V22 þ p2

(7.32)

cp T 1 þ q þ

V12 V2 ¼ cp T2 þ 2 2 2

(7.33)

where q is the energy added (or chemical enthalpy) per unit mass and V1 ¼ Vc . Because the equation of state p1 ¼ ρ1 RT1 relates known variables, the only state equation is, assuming a constant gas constant, p2 ¼ ρ2 RT2

(7.34)

There are four equations, Eqns. (7.31) through (7.34), for five unknown variables: V1 , V2 , p2 , T2 , and ρ2 . Therefore, there exist an infinite number of solutions to these four equations. Physical reasoning dictates that there must exist an intrinsic variable (an eigenvalue) in the mixture system, which is expected to be the wave propagation speed that is a function of the variables of the unburned gas. It is desirable to develop a relationship to demonstrate the existence of this unique variable before attempting to obtain solutions of these four equations. Similar processes in Chapter 4 leading to the Rankine-Hugoniot relation can be followed. In the following, R, cp , and γ are assumed to be constant throughout the entire flow field (i.e., in both the burned and unburned regions). Combining Eqns. (7.31) and (7.32), one finds the following two expressions:

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198

7 One-Dimensional Flows with Heat Transfer

V12

1 ¼ 2 ρ1

p2  p1 1 1 ρ ρ

1 ¼ 2 ρ2

p2  p1 1 1 ρ ρ

1

!

or V12 ¼ –v21

2

p2  p1 –v1  –v2

(7.35)

and V22

1

!

or V22 ¼ –v22

2

p2  p1 –v1  –v2

(7.36)

Dividing through these two expression, respectively, by their speed of sound (a2 ¼ γRT ¼ γP=ρ) leads to p2 =p1  1 (7.37) γM12 ¼ 1  ρ1 =ρ2 and γM22 ¼

1  p1 =p2 ρ2 =ρ1  1

(7.38)

Both Equations (7.37) and (7.38) suggest that solutions to Eqns. (7.31) through (7.34) only exist for either (i) p2 =p1 > 1 and ρ2 =ρ1 > 1 (i.e., compression; these waves are called detonation waves) or (ii) p2 =p1 < 1 and ρ2 =ρ1 < 1 (i.e., expansion; these waves are called deflagration waves). Recalling that cp ¼ γR=ðγ  1Þ, the energy equation, Eqn. (7.33) can be written as  γ 1 RðT2  T1 Þ  V12  V22 ¼ q γ1 2 Substituting Eqns. (7.35) and (7.36) into this expression leads to



γ p2 p1 1 1 1  þ  ðp2  p1 Þ ¼q γ  1 T2 T1 2 ρ1 ρ2 Multiplying both sides of Eqn. (7.39) by ρ2 and rearranging terms leads to
 γþ1 2 q ρ2 γ1 þ RT1 ρ1  1 p2
 ¼ γþ1 p1  ρ2 γ1

(7.39)

(7.40a)

ρ1

or alternatively,
p2 ¼ p1

 2q þ RT  vv––21
 1 γþ1 v–2 γ1 v–1  1

γþ1 γ1

(7.40b)

It is readily seen that for q ¼ 0 these two equations reduce to the Rankine-Hugoniot (or simply, Hugoniot) relationship for shock waves derived in Chapter 4, Eqn. (4.72). The detailed derivation of Eqn. (7.40) is left as an exercise (i.e., Problem 7.10). In addition to the thermodynamic quantities contained in the shock adiabat

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7.6 Detonation and Deflagration Waves

199

described in Chapter 4, Eqns. (7.40a) and (7.40b) contains the heat release per unit mass. The function p2 ¼ p2 ðρ2 ; qÞ or p2 ¼ p2 ð–v2 ; qÞ is called the detonation adiabat and will be discussed in more detail in the following. p2 p1 D, J (I)

B (II) Not possible

O

1

A

(III)

C

(IV) q > o; Eqn. (7.40 b)

q = o; Eqn. (4.72) 0 γ –1 γ +1

υ2 υ1

Figure 7.8 The Hugoniot curve with δq > 0 plotted using Eqn. (7.40b) Physical reasoning leads to two steady-state solutions: Chapman-Jouget detonation wave (point J) and deflagration wave (regimes III).

The Hugoniot relationship of p2 =p1 vs. –v2 =–v1 is plotted using Eqn. (7.40b) (see Fig. 7.8). It can be seen that the pressure increase due to chemical reaction is higher than that without heat addition; Eqn. (4.72) is also seen in Fig. 7.8 for comparison. A different Hugoniot curve has to be drawn for a different value of q. The propagating wave causes the burned gas to move with a speed Vb ¼ Vs  V2 ¼ V1  V2 . Using Eqns. (7.35) and (7.36) yields pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v1  –v2 Þðp2  p1 Þ (7.41) V b ¼ ð– The discussion above pertaining to Eqns. (7.38) and (7.39) excludes portion of the Rankine-Hugoniot curve, i.e., region II between points A and B in Fig. 7.7, as possible solutions. There are two points on the curve that require particular attention – tangency points C and J, called Chapmen-Jouget points. Points C and J are the intersections between the curve and the tangent lines drawn from the initial point (point I). For these two points, an angle, αJ , is defined as

ðp2  p1 Þ 1 1 tan αJ ≡ or tan αJ ≡  ðp2  p1 Þ (7.42) v1  –v2 Þ ρ1 ρ2 ð–

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200

7 One-Dimensional Flows with Heat Transfer

Therefore, Eqn. (7.35) can be rewritten as pffiffiffiffiffiffiffiffiffiffiffiffiffi v1 tan αJ V1 ¼ –

(7.43)

Along the Hugoniot curve, Tds2 ¼ de2 þ p2 dð1=ρ2 Þ

(7.44)

It is desirable to express e2 in terms of p and –v(or ρ), which has been used so far. The energy equation, Eqn. (7.33), can be rewritten as ht1 þ

V12 V2 ¼ ht2 þ 2 2 2

where ht1 and ht2 equal the total energy, that is, the sum of thermal energy and chemical (rather than kinetic) energy of the fresh and burned gases, respectively, such that h t ¼ h þ ho ¼ c p T þ h o where ho is the baseline reference enthalpy in the standard state of gas species, and q ¼ ho1  ho2 Equation (7.33) then becomes h1 þ ho1 þ

V12 V2 ¼ h2 þ ho2 þ q þ 2 2 2

(7.45)

Therefore h2  h1 ¼

V12 V22  2 2

Substituting part of the result from Eqn. (7.39):

V12 V22 1 1 1  ¼ ðp2  p1 Þ þ 2 ρ1 ρ2 2 2

(7.46)

and h ¼ e þ ð p=ρÞ into Eqn. (7.44) yields



1 1 1 e2  e1 ¼ ðp2 þ p1 Þ  2 ρ1 ρ2

(7.47)

Differentiating the above expression, while noting that state 1 is given, leads to



1 1 1 1 1 de2 ¼  ðp2 þ p1 Þd  þ dp2 2 ρ2 2 ρ1 ρ2 With the above expression, Eqn. (7.44) becomes



1 1 1 1 1 Tds2 ¼ ðp2  p1 Þd  þ dp2 2 ρ2 2 ρ1 ρ2

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7.6 Detonation and Deflagration Waves

201

Therefore, along the Hugoniot curve, 2 3

ds2 1 1 1 4ðp2  p1 Þ dp2 5
þ
  T
¼ 1 1 2 ρ1 ρ2 d ρ1 d ρ1 ρ ρ 2

1

2

(7.48)

2

Because region II has been eliminated from possible solutions, then in regions I, III, and IV ðp2  p1 Þ
 >0 1 1 ρ ρ 1

2

At tangency points C and J, the above inequality holds. Since at points C and J, dp2 ðp  p 1 Þ
¼
2  1 1 1 d ρ  ρ ρ 2

1

(7.49)

2

one can conclude that 2

3 ds 4
25 d ρ1 2

¼0

(7.50)

K; J

Differentiating both sides of Eqn. (7.48) with respective to 1=ρ2 for constant temperature leads to

2 d 2 s2 1 1 1 d p2  (7.51)
2 ¼
2 2 T ρ ρ 1 2 d 1 d ρ1 ρ 2

2

The derivative of p2 in the above equation can be obtained from Eqn. (7.40a). After some differential and algebraic manipulations (this is left as an end-of-chapter exercise, Problem 7.10), 8
h

i9 γþ1 γþ1 γþ1 2q >

2 > = < 2 þ 1  γ1 γ1 γ1 RT1 d s2 p1 ρ2 ρ2 (7.52) ¼   1
2
3 > T ρ1 ρ1 > γþ1 ; : d 1  ρ2 γ1

ρ2

ρ1

Some useful observations of Eqn. (7.52) can be made as follows – (1) The numerator in the bracelet of Eqn. (7.52) is always < 0 for q ≥ 0. (2) In region of III and IV of Fig. 7.8 (expansion wave), which includes point C, p2 < p1 and ρ2 =ρ1 < 1 making d2 s2 =dð1=ρ2 Þ2 < 0. Therefore, 2 3 2 6 d s2 7 4
2 5 d ρ1 2

p1 and ρ2 =ρ1 > 1, the displacement of the detonation adiabat above that of the shock adiabat gives the result v2 =– – v1 > ðγ  1Þ=ðγ þ 1Þ or ρ2 =ρ1 < ðγ þ 1Þ=ðγ  1Þ. Therefore, 2 3 2 6 d s2 7 4
2 5 > 0 d ρ1 2

(7.53b)

I

As suggested by Eqn. (7.50), the processes near points C and J are essentially isentropic. Therefore the local speed of sound can be determined as 2 3

∂p 1 ∂p 2 2 ¼  2 4
5 (7.54) a22 ¼ ∂ρ2 s ρ2 ∂ 1 ρ2

s

By substituting Eqn. (7.49) for point C and J and equating the result with Eqn. (7.36), 2 3  2 1 4 ∂p2 5 1 ðp2  p1 Þ  ¼ V22 (7.55) a2 K;J ¼  2
 ¼ 2
1 ρ2 ∂ ρ2 1  1 ρ2

s

ρ1

ρ2

That is, at points C and J, the burned gas has a sonic velocity relative to the wave: M2;J ¼ M2;C ¼ 1

(7.66)

Because V1 > V2;C and T2 > T1;C (and a2 > aC ), M1;J > 1. In fact the compression wave in region I has a Mach number ðM1 ÞI > 1. Therefore, the wave in this region is a shock wave supported by the chemical heat-releasing reaction. So far the analysis of Eqns. (7.31) through (7.34) has produced the possible solutions represented by region I, III, and IV, including point J. In the following, physical reasoning will help to further eliminate some of these solutions. Region I above point J is the strong detonation region because ðp2 =p1 Þ > ðp2 =p1 ÞJ . Because of the higher compression ratio, T2 > T2;J and V2 < V2;J and M2 < M2;J ¼ 1. (The fact M2 < M2;J is also consistent with entropy being minimum at J.) In this region, the subsonic downstream condition enables rarefaction and dissipative effects to overtake and weaken the shock wave, and move down the Hugoniot curve toward point J. Because M2;J ¼ 1, downstream effects cannot reach the shock wave. For the portion of the region between B and J, ðp2 =p1 Þ < ðp2 =p1 ÞJ , T2 > T2;J , and M2 > M2;J ¼ 1; it is the weak detonation region. However, the flow immediately downstream of the shock is subsonic and heat addition in a one-dimensional flow on a Rayleigh line cannot be accelerated past the sonic condition. Therefore, for region I, point J is the only possible solution. Thus, conditions represented by J correspond to a self-sustaining steady-state detonation wave, where a supersonic wave is supported by a heat-releasing reaction and leaves the burned gas traveling at a sonic speed relative to the wave.

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7.6 Detonation and Deflagration Waves

203

In region IV (i.e., to the right of point C), p2 is only slightly less than p1 , and ρ2 is much less than ρ1 (and V2 > V1;K ), representing more expansion and acceleration downstream of the expansion than those at K. The wave is an expansion wave and is thus subsonic. Due to this excessive expansion it is the strong deflagration region, with T2 < T1;K and a2 < aK , leading to M2 > M1;K ¼ 1. It is not possible to accelerate a subsonic flow to supersonic speed by heat addition in a constant area duct. Region IV is thus not physically possible. Region III represents the solution where a subsonic expansion wave is supported by heat release that leaves the burned gas accelerating to a higher speed than the unburned gas, relative to the wave. This region is called the weak deflagration region where ðp2 =p1 Þ < ðp2 =p1 ÞC and ρ2 < ρC . As a consequence, the steady-state solution on the Hugoniot curve of Fig. 7.8 consists of point J and region III. The wave propagation corresponding to point J is called the Chapman-Jouquet (or simply C-J) detonation wave. Region I above point J, although not a steady-state solution, is possible during the transient period before it quickly resettles to J due to external weakening effects. Region III comprises solutions for subsonic propagation of pressure and combustion waves, called deflagration waves. It is of interest to determine the unique detonation speed of the steadily propagating C-J detonation wave. Assuming that γ and R remain across the wave, the continuity, momentum, and energy equations, Eqn. (7.31) through (7.33), can be rewritten as rffiffiffiffiffiffi p2 M1 T2 ¼ (7.67) p1 M2 T1 p2 1 þ γM12 ¼ p1 1 þ γM22

(7.77)

q cp

(7.78)

Tt2 ¼ Tt1 þ

where Tt ¼ h þ V 2 =2 is the stagnation temperature. At J, the isentropic relationship T ¼ Tt =½1 þ ðγ  1ÞM2 =2 applies, as permitted by Eqn. (7.50). Combining Eqns. (7.67) and (7.77) leads to  2 T2 M22 1 þ γM12 ¼   T1 M2 1 þ γM2 2 2 1 #  2 #" Þ M22 1 þ γM12 1 þ ðγ1 M22 2 h i¼ 1þ   Þ Þ 2 2 1 þ γM2 2 2 1 þ ðγ1 M M cp T1 1 þ ðγ1 2 M1 2 1 1 2 q

(7.79)

"

(7.80)

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204

7 One-Dimensional Flows with Heat Transfer

For C-J detonation (M2 ¼ 1), with the chemical heat release term on the left-hand side of Eqn. (7.81) is significantly larger than unity, " # 2 #" ðγþ1Þ 1 þ γM12 q 2 h i¼ (7.81) Þ Þ 2 2 M12 ð1 þ γÞ2 1 þ ðγ1 cp T1 1 þ ðγ1 2 M1 2 M1 The steady C-J detonation wave travels at a Mach number ranging from approximately 5 and 10 for most stoichiometric fuel-oxygen mixtures. Therefore, with the assumption of M12 ¼ 1, Eqn. (7.81) becomes q γ2 ≈ cp Tt1 ðγ þ 1Þðγ  1Þ

(7.82)

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q 2ð γ þ 1Þ cp T1

(7.83)

and M1 ≈

However, examination of Eqn. (7.81) suggests that the largest error associated with the M12 ¼ 1 assumption might very well arise from the term ½1 þ ðγ  1ÞM12 =2; as for M1 ¼ 5 and γ ¼ 1:4, ðγ  1ÞM12 =2 ¼ 5, which is not much larger than 1. A rearrangement by eliminating the ½1 þ ðγ  1ÞM12 =2 term from both sides of Eqn. (7.81) leads to rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 q 2ð γ þ 1Þ (7.84) M1 ≈ γ cp T1 This value of M1 is smaller than that of Eqn. (7.83) by a factor of 1=γ. Calculate the C-J detonation wave Mach number of the stoichiometric hydrogen-oxygen mixture, initially at 1 atmosphere and 300 K. Also (i) find the speed of the burned gas relative to the laboratory and the pressure downstream of the wave. The heat capacity of the fresh mixture and the water vapor is c p ¼ 2:4 kJ=kg ⋅ K and the heat from the reaction is 119,960 kJ/kgH2. Assume the product is water vapor (γ ¼ 1:327 and R ¼ 0:461 kJ=kg ⋅ K and cp ¼ 2:2 kJ=kg ⋅ K); (ii) the density ratio using the M2 ¼ 1 assumption and the Hugoniot relationship, Eqn. (7.40a). EXAMPLE 7.8

Solutions – The value R of the fresh mixture is (H2 þ 12 O2 ), which has an average molecular weight of 12 kg/kmol and thus 13,329 kJ per kilogram of the mixture. Therefore, R ¼ ð8:314 kJ=kmol ⋅ KÞ=ð12 kg=kmolÞ ¼ 0:6928 kJ=kg ⋅ K. rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 q 1 13; 329 2ðγ þ 1Þ  2  2:4  ¼ 6:73 M1 ¼ ¼ γ cp T1 1:4 2:4  300

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7.6 Detonation and Deflagration Waves

205

p2 1 þ γM12 1 þ 1:4  6:6042 ¼ 25:86 ¼ ¼ p1 1 þ γM22 1 þ 1:4  2  2 1 þ 1:4  6:732 T2 M22 1 þ γM12 ¼ ¼ 15:90   ¼ T1 M2 1 þ γM2 2 6:732  ð1 þ 1:4Þ2 2 1 Assuming constant γ and R, ρ2 p2 T1 ¼  ¼ 1:63 ρ1 p1 T2 V 2 ¼ M2

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi γ2 R2 T2 ¼ 1  1:327  461  15:90  300 ¼ 1; 708:34 m=s

V1 ¼ M1

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi γ1 R1 T1 ¼ 6:73  1:4  693  300 ¼ 3; 630:8 m=s

For a moving detonation wave, the speed of burned gas relative to the laboratory is Vb ¼ V1  V2 ¼ 1; 922:5 m=s Vb 1; 922:5 Mb ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ≈ 1:125 γ2 R2 T2 1:327  461  15:90  300 (The burned gas moves at supersonic speed to a laboratory observer, although at a sonic speed relative to the wave.) Comments – As comparisons, a shock wave with M1 ¼ 6:73 and γ ¼ 1:4 without chemical heat release causes p2 =p1 ≈ 52 and T2 =T1 ≈ 8:5. The reason the detonation wave produces a higher value is easy to understand: the chemical reaction adds energy to the gas. The lower p2 =p1 by detonation might be attributed to the extra increase in entropy and thus smaller pt2 ; coupled with the sonic condition, the □ equation yields a lower value. Another way for determining V1 is to take advantage of the fact that M2 ¼ 1 and that continuity requires that V1 ¼ ðρ2 =ρ1 ÞV2 . ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

v u ρ2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ρ2 u Tt2 ρ 2γ2 R2 tγ2 R2 V1 ¼ Tt2 ¼ 2 γ2 R2 T2 ¼ ρ1 ρ1 ρ1 γ2 þ 1 1 þ γ2 12 2M2

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ρ 2γ2 R2 q Tt1 þ ¼ 2 cp;2 ρ1 γ2 þ 1

(7.85)

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206

7 One-Dimensional Flows with Heat Transfer

Equation (7.85) requires first determining ρ2 =ρ1 . This density ratio is close to 1.8 for many fuels under stoichiometric conditions (Glassman and Yetter, 2008). EXAMPLE 7.9 Use Eqn. (7.85) to determine the detonation wave speed of the mixture of Example 7.8. sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ρ 2γ2 R2 q Tt1 þ V1 ¼ 2 cp;2 ρ1 γ2 þ 1

Since the determination of Tt1 requires knowledge of M1 , which is itself to be determined, it is appropriate to use sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ρ 2γ2 R2 V1 ¼ 2 Tt2 ρ1 γ2 þ 1

Tt2 ¼

1  cp;2

 1 ð2:4  300 þ 13; 329Þ ≈ 6; 386K cp;1 T1 þ q ¼ 2:2

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2  1:327  461  6; 386 V1 ¼ 1:8  ≈ 3; 300 m=s 2:327 V1 3; 300 M1 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ≈ 6:12 γ1 R1 T1 1:4  692:8  300 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi V2 ¼ M2 γ2 R2 T2 ¼ 1 

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6; 368 1:327  461  ≈ 1; 830 m=s 1:3271 1þ 2 1

Comments – The results of Examples 7.8 and 7.9 agree within 10% for V1 and M1 , and approximately 7% for V2 . □

Problems Problem 7.1 Derive Eqn. (7.15d). Problem 7.2 Show the steps leading to Eqn. (7.24). Problem 7.3 An air flow enters a frictionless constant-area duct with M1 ¼ 0:3, T1 ¼ 258 C, and ρ1 ¼ 2:20 kg=m3 and exits the duct with M2 ¼ 0:7. Find the heat transfer and changes in entropy and stagnation pressure between the duct entrance and exit.

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Problems

207

Problem 7.4 An air flow enters a frictionless constant-area duct with M1 ¼ 2:0 and exits the duct with M2 ¼ 2:5, T2 ¼ 300 K, and p2 ¼ 101:3 kPa. Find the heat transfer and changes in entropy and stagnation pressure between the duct entrance and exit. Problem 7.5 A circular pipe (5 cm in diameter) is connected to an air reservoir (at 700 kPa and 450 K) through a converging nozzle. What is the maximum flow rate that can be delivered by this nozzle-pipe assembly if the cooling of the air (i.e., heat loss) is 50 kJ/kg and the back pressure is 0 kPa (i.e., vacuum)? Determine the pressure and the Mach number at the exit plane under these conditions? Problem 7.6 A circular pipe (5 cm in diameter) is connected to an air reservoir (at 700 kPa and 450 K) through a converging nozzle. The cooling rate of the air is 50 kJ/ kg. Determine the back pressure when a normal shock stands at the exit. Problem 7.7 A long constant-area frictionless duct is attached to an air reservoir (at 300 kPa and 300 K) through a converging-diverging nozzle with smooth geometrical transition in shapes. Assume the area ratio of the nozzle exit to the throat is 3. Assuming there is no shock wave in the duct-nozzle assembly, find the amount of heat transfer for the duct exit condition to be sonic. Also find the exit plane pressure, the entropy change, and pressure loss. Assume that cp ¼ 1:004 kJ=kg ⋅ K throughout the entire flow. Problem 7.8 A long frictionless circular duct, having a diameter of 2 cm, is connected to an air reservoir (at 500 kPa and 400 K) by a converging nozzle. It is to deliver air into another reservoir that is a vacuum. Determine the mass flow rate and the change in stagnation pressure if (1) the heat is added to the air through the duct wall at 50.2 kJ/kg, and (2) if heat is taken away from the air (i.e., cooling) at 50.2 kJ/ kg. Assume that cp ¼ 1:004 kJ=kg ⋅ K throughout the entire flow. Problem 7.9 A frictionless circular duct is connected to an air reservoir (at 300 kPa and 300 K) through a converging-diverging nozzle. The duct inlet to nozzle throat area ratio is 2.0 and the heat addition to the flow through the duct wall is 50 kJ=kg. Assume that cp ¼ 1:004 kJ=kg ⋅ K throughout the entire flow. (1) What is the maximum back pressure to avoid a normal shock through the duct-nozzle device? Under this condition, what is the duct exit velocity? (2) What would be the combination of the amount of heat addition and back pressure for the duct exit condition to be sonic? Problem 7.10 Derive in details Eqn. (7.40). Problem 7.11 (a) Use the differential forms of mass, momentum, and the ideal gas law to show that  dTt dV  1  M2 ¼ V T

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208

7 One-Dimensional Flows with Heat Transfer

That is, the Rayleigh line flows heat addition/removal in the subsonic regime, leading to flow acceleration/deceleration, and to deceleration/acceleration in the supersonic regime. (b) Use the differential forms of mass, momentum, and energy equations, combined with results in Part (a) to show that for a Rayleigh line

dT 1  γM2 dTt ¼ T 1  M2 T and that the maximum temperature occurs at M ¼ sonic branch.

pffiffiffiffiffiffiffi 1=γ, that is, on the sub-

Problem 7.12 Show all steps leading to Eqn. (7.52). Problem 7.13 The fresh reactant mixture of Example 7.8 is now diluted with nitrogen or helium gases, so that it is (H2 þ 12 O2 þ 5N2 ) and (H2 þ 12 O2 þ 5He). Find the detonation wave speeds and pressure and temperature ratios of these two mixtures. Compare with the results of Example 7.8, especially those of V1 and M1 , and try to draw conclusions. Problem 7.14 The following expression for pressure ratio applies to both detonation and shock waves: p2 1 þ γM12 ¼ p1 1 þ γM22

Eqn. (4.18a)

(1) Assuming that γ2 ¼ γ1 ¼ γ across the C-J detonation wave, compare the magnitudes of pressure ratios of the detonation and the shock waves with the same M1 . (2) Find the difference between the pressure ratios of these two waves. (3) For the same very large M1 , show that the pressure ratio across the shock is approximately twice that across the detonation wave. Problem 7.15 Assume that a jet combustor is a constant-area duct where viscous effects are negligible. If the combustion leads to a stagnation temperature equal to 4.5 and the desired Mach number of 0.9, determine (a) the duct inlet Mach number, and (b) the amount of heat added. Foe convenience, assume and throughout the combustion and flow processes and an inlet temperature of 250 K.

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8

Equations of Multidimensional Frictionless Flow Subject to Small Perturbation

The one-dimensional theory discussed in previous chapters demonstrates essential features of gas dynamics signaling mechanism, compression, expansion, shock wave and its formation, flow turning, effects of area changes, friction, and heat transfer, with applications to lift and drag on airfoils and channel flows. In general, there are effects of multidimensional and unsteady nature, such as curved shocks, shock, and expansion wave propagation. While simplifications to one-dimensionality may serve as good approximations, many practical flows are inherently multidimensional and unsteady. To focus on the multidimensional and unsteady effects, this chapter will explore inviscid flows without heat transfer.

8.1 Differential Control Volume Approach for Mass Conservation The conservation of mass (continuity) equation for an arbitrary control volume is given in Chapter by Eqn. (2.4): ð ð ð ð ∂ ~ or ∂ ~ ~ ⋅ dAÞ ~ ⋅ dAÞ ρd–V þ ðρV ρd–V ¼  ðρV (2.4) 0¼ ∂t CV– ∂t CV– CS CS Consider the differential control volume having an infinitesimal size shown in Fig. 8.1, at whose center all the fluid properties are defined. The time rate of change within the control volume (or CV) can be rewritten as ð ∂ ∂ρ dxdydz ρd– V¼ ∂t CV– ∂t ~ ¼ ~iu þ~jv þ ~ Let the velocity field be represented by the vector, V kw. The mass flux term across the boundary of the boundary (i.e., the control surface, or CS) in Eqn. (2.4) becomes 



 

ð ∂ðρuÞ dx ∂ðρuÞ dx ∂ðρvÞ dy ~ ¼ ~ ⋅ dAÞ ðρV ρu þ  ρu  dydz þ ρv þ ∂x 2 ∂x 2 ∂y 2 CS

 



 ∂ðρvÞ dy ∂ðρwÞ dz ∂ðρwÞ dz  pv  dxdz þ ρw þ  ρw  dxdy ∂y 2 ∂z 2 ∂z 2

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210

8 Equations of Multidimensional Frictionless Flow Pathline z u3 (or w)

Control volume, (dx dy dz)

g Pathline dz u2 (or u) Pathline u1 (or u)

dx dy

Pathline

y

x Figure 8.1 The differential control volume in the Cartesian coordinate system.

Therefore,   ∂ðρuÞ ∂ðρvÞ ∂ðρwÞ ~ ~ þ þ ðρV ⋅ dAÞ ¼ dxdydz ∂x ∂y ∂z CS

ð

Thus the mass conservation for the differential control volume becomes ∂ρ ∂ðρuÞ ∂ðρvÞ ∂ðρwÞ þ þ þ ¼0 ∂t ∂x ∂y ∂z

(8.1a)

∂ ∂ ∂ or, by using the divergence operator r ≡~ı ∂x þ~j ∂y þ~ k ∂z ,

∂ρ ~Þ ¼ 0 þ r ⋅ ðρV ∂t

(8.1b)

The continuity equation can also be rewritten by using the substantial, or Eulerian, differential operator

D ∂ ~ ∂ ∂ ∂ ∂ ≡ þ V ⋅r ¼ þ u þ v þ w (8.1c) Dt ∂t ∂t ∂x ∂y ∂z ~ ⋅r where the first term on the right-hand side accounts for unsteadiness and V represents the convective effects or the changes a fluid particle experiences due to convection. Thus, the Eulerian derivative represents changes in fluid particles due to the combined unsteady and convective effects. Equation (8.1b) then becomes ∂ρ ~ ~ ¼0 þ V ⋅ rρ þ ρr ⋅ V ∂t

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8.2 Conservation of Momentum

211

and therefore Dρ ~ ¼0 þ ρr ⋅ V Dt

(8.1d)

~ ¼ V ~ ⋅ rρ so that Dρ=Dt ¼ It is noted that for steady flows, ∂ρ=∂t ¼ 0, ρr ⋅ V ~ V ⋅ rρ ≠ 0 in general except for incompressible flows where ρ ¼ constant. For steady flows, Eqn. (8.1b) reduces to ~Þ ¼ 0 r ⋅ ðρV

(8.1e)

If the flow is incompressible, ρ remains unchanged for time and location, and both Eqns. (8.1b) and (8.1d) become ~¼0 r⋅V

(8.1f)

In cylindrical coordinates, the del operator, ∇, is expressed as r ¼~ er

∂ 1 ∂Vr ~ ∂ þ~ eθ þk ∂r r ∂θ ∂z

(8.1e) (8.2)??

where~ e r ,~ e θ , and ~ k are unit vectors in the radial, circumferential, and axial directions, respectively.

8.2 Conservation of Momentum The control volume form of momentum conservation has been given in Chapter 2 as ð ð X  ∂ ~ ~ ~ ~ ðρV ~ ⋅ dAÞ ¼ ρV d–V þ F V (2.5) i i system ∂t CV– CS Due to the vector nature of momentum, it is convenient and instructive to first obtain the result in the x-direction for the differential CV shown in Fig. 8.1, as follows. ð ∂ ∂ðρuÞ dxdydz ρud– V¼ ∂t CV– ∂t ð CS

~Þ¼ ~ ⋅ dA uðρV







∂u dx ∂ðρuÞ dx ∂u dx ρu þ  u uþ ∂x 2 ∂x 2 ∂x 2

 



∂ðρuÞ dx ∂u dy ∂ðρvÞ dy ρu  dydz þ u þ ρv þ ∂x 2 ∂y 2 ∂y 2



 

∂u dy ∂ðρvÞ dy ∂u dz  u ρv  dxdz þ u þ ∂y 2 ∂y 2 ∂z 2





 ∂ðρwÞ dz ∂u dz ∂ðρwÞ dz ρw þ  u ρw  dxdy ∂z 2 ∂z 2 ∂z 2

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212

8 Equations of Multidimensional Frictionless Flow

For frictionless flow, the forces acting on the fluid particles can be decomposed into two groups: one due to pressure ( p) and the other due to all body forces (~ f , including gravitation and electromagnetic forces, etc.). Therefore, in the x-direction this becomes



X  ∂p dx ∂p dx ~ ¼ p dydz  p þ dydz þ ρfx dxdydz F i i system ∂x 2 ∂x 2 ¼

∂p dxdydz þ ρfx dxdydz ∂x

Combining the above three results while neglecting the second- and higher-order terms, the differential equation for conservation in the x-direction is 

∂p ∂ðρuÞ ∂ðρuÞ ∂ðρvÞ ∂ðρwÞ ∂u ∂u ∂u þ ρfx ¼ þu þu þu þ ρu þ ρv þ ρw ∂x ∂t ∂x ∂y ∂z ∂x ∂y ∂z

Substituting the continuity equation, Eqn. (8.1a), and dividing through the x-momentum equation by ρ yields

1 ∂p ∂u ∂u ∂u ∂u  þ fx ¼ þþ u þv þw (8.2a) ρ ∂x ∂t ∂x ∂y ∂z Similarly, the y- and z-momentum equations can be found, respectively, as

1 ∂p ∂v ∂v ∂v ∂v þ fy ¼ þ u þv þw  ρ ∂y ∂t ∂x ∂y ∂z





1 ∂p ∂w ∂w ∂w ∂w þ fz ¼ þ u þv þw ρ ∂z ∂t ∂x ∂y ∂z

(8.2b)

(8.2c)

In vector notation, these three components of the momentum equation can be combined as ~ ∂V ~ DV ~ ⋅ rÞV ~ ¼  1 rp þ ~ ¼ þ ðV f Dt ∂t ρ

(8.2d)

If the frictional and gravitational forces can be neglected, Eqn. (8.2d) is reduced to ~ DV 1 ¼  rp Dt ρ

(8.2e)

Equation (8.2e) is the well-known Euler’s equation for inviscid flow without body forces.

8.3 Conservation of Energy The principle of conservation of energy, Eqn. (2.19) for a frictionless flow with no work done by external mechanisms can be rewritten as

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8.3 Conservation of Energy

213




ð  1 1 ~ ~ ⋅ dA e þ V 2 ρd–V þ h þ V 2 ρV (8.3) 2 2 CV CS   
~ ⋅V ~ =2 (with where e denotes internal energy and ht ¼ h þ V 2 =2 ¼ h þ V ~ ⋅V ~ =2) is the stagnation enthalpy. For the control V 2 =2 ¼ ðu2 þ v2 þ w2 Þ=2 ¼ V volume shown in Fig. 8.1,
 h
i _ ¼ ρqdxdydz ~ dxdydz ¼ ρq_ þ ρ ~ ~ dxdydz _ Q_  W þρ ~ f ⋅V f ⋅V (8.4a) _ ¼ ∂ Q_  W ∂t

ð

where q_ is the heat transfer rate per unit mass of the gas. Following the similar procedure for mass conservation, the right-hand side terms of Eqn. (8.3) can be rewritten as follows. For  brevity and convenience in derivation, let  
~ ⋅V ~ =2 . Then é ¼ e þ V 2 =2 ¼ e þ V ∂ ∂t ð CS



1 ∂ðρéÞ dxdydz e þ V 2 ρd–V ¼ 2 ∂t CV

ð

(8.4b)


 1 ~ ~ ⋅ dA h þ V 2 ρV 2 







 ∂ht dx ∂ðρuÞ dx ∂ht dx ∂ðρuÞ dx ¼ ht þ ρu þ  ht  ρu  dydz ∂x 2 ∂x 2  ∂x 2 ∂x 2  ∂ht dy ∂ðρvÞ dy ∂ht dy ∂ðρvÞ dy ρv þ  ht  ρv  dxdz þ ht þ 2 ∂y 2 2 ∂y ∂y 





∂y 2

 ∂ht dz ∂ðρwÞ dz ∂ht dz ∂ðρwÞ dz ρw þ  ht  ρw  þ ht þ ∂z 2 ∂z 2 ∂z 2 ∂z 2 (8.4c)

Expanding and neglecting the second- and higher-order terms of Eqn. (8.4c), and then combining it with Eqns. (8.3a) and (8.3b) leads to ~¼ ρq_ þ ρ~ f ⋅V


 ∂ðρéÞ ∂ðρuht Þ ∂ðρvht Þ ∂ðρwhtÞ ∂ðρéÞ ~ þ þ þ þ r ⋅ ρht V ¼ ∂t ∂x ∂y ∂t ∂z

(8.5a)

or



 ~ ¼ ∂ ρe þ 1 ρV 2 þ r ⋅ ρ h þ 1 V 2 V ~ ρq_ þ ρ~ f ⋅V ∂t 2 2

(8.5b)

Because h ¼ e þ p=ρ; 

 


 ∂ 1 2 1 2 ~ ~ ~ ~ _ ρ eþ V ρq þ ρf ⋅ V ¼ þ r ⋅ ρ e þ V V þ r ⋅ pV ∂t 2 2

(8.5c)

The first two bracketed terms on the right-hand side of Eqn. (8.5c) represent, respectively, the rate of storage of the combined internal and kinetic energy within

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8 Equations of Multidimensional Frictionless Flow

the
CV and rate of convection of the sum outside of the CV. The third term, ~ , is the net “flow work” per unit volume that is needed to push the fluid r ⋅ pV particle against the pressure so as to leave the CV. This equation can also be rewritten as   
 1 ∂ρ
 ∂e ∂ρ 2 ~ ~ ~ ~ ~ þ r ⋅ ρV þ V þ r ⋅ ρV ρq_ þ ρf ⋅ V ¼ ρ þ ρV ⋅ re þ e ∂t ∂t 2 ∂t 




 ∂ 1 2 ~ ⋅ r 1 V2 ~ þ ρ V þ ρV þ r ⋅ pV (8.5d) ∂t 2 2 The first bracket in the above expression becomes De=Dt. With the help of continuity, Eqn. (8.1b), the second and third brackets on the right-hand side are identically zero and the fourth bracket becomes " # "



# ~ ~ ~ ∂ V 1 ∂ V 1 ∂ V 2 2 ~ ⋅ r V ¼ ρV ~ ⋅ rV ~ ~⋅ ~⋅ ~⋅ þ ρV þr V þV ρV ¼ ρV ∂t 2 ∂t 2 ∂t ~ ~ ⋅ DV ¼ ρV Dt Equation (8.5c) then becomes
 ~ ~ ⋅ D V þ r ⋅ pV ~ ~ ¼ ρ De þ ρV ρq_ þ ρ~ f ⋅V Dt Dt

(8.5e)

Neglecting ~ f and assuming the change in elevation is small, the momentum equation, ~ =Dt ¼ rp=ρ, which can be substituted into Eqn. Eqn. (8.2e), dictates that DV (8.5e) to yield

De rp ~ ~ þV ~ ⋅ rp þ ρV ⋅  ρq_ ¼ ρ þ pr ⋅ V (8.5f) Dt ρ Therefore ρq_ ¼ ρ

De ~ þ pr ⋅ V Dt

(8.5g)

Therefore the conservation of energy indicates that the rate of heat transfer to the CV is equal to the sum of the increase in the internal energy of the gas passing ~ > 0), or on the gas through the CV and the work done by the gas expansion (if r ⋅ V ~ < 0). (i.e., compression and r ⋅ V To express the energy equation in terms of stagnation enthalpy, ht , Eqn. (8.5c) can be rewritten by incorporating a pressure term, as 

 



 ∂p ∂ p 1 2 p 1 2 ~ ~ ~  ρ eþ þ V  r ⋅ pV ρq_ ¼ þ r ⋅ ρ e þ þ V V þ r ⋅ pV ∂t ρ 2 ρ 2 ∂t (8.5h) With the help of continuity, Eqn. (8.1b), this expression simplifies to

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8.3 Conservation of Energy

ρq_ ¼ ρ

Dht ∂p  ∂t Dt

215

(8.5i)

It is also possible to express the energy conservation equation in terms of h ð¼ e þ p=ρÞ: De Dh Dð p=ρÞ Dh 1 Dp p Dρ ¼  ¼  þ Dt Dt Dt Dt ρ Dt ρ2 Dt ~ so that the above expression The continuity, Eqn. (8.1d), provides Dρ=Dt ¼ ρr ⋅ V becomes De Dh 1 Dp p ~ ¼   r⋅V Dt Dt ρ Dt ρ Substituting this expression for De=Dt into Eqn. (8.5g) yields ρq_ ¼ ρ

Dh Dp  Dt Dt

(8.5j)

For steady, adiabatic flows neglecting the effects of friction and body/gravitational forces, Dht =Dt ¼ 0. That is, the stagnation enthalpy of a gas particle remains constant along its path. In steady flows, the path line coincides with the streamline. Therefore under these conditions, ht ¼ h þ

V2 p V2 ¼eþ þ ¼ constant ρ 2 2

(8.5k)

along a streamline. The constant in Eqn. (8.5k) in general varies from one to the other streamline, as expected in multidimensional flows. This result is the same as that for isentropic flows discussed in Chapter 4. It is noteworthy that Eqn. (8.5j) can be derived alternatively by starting with Eqn. (2.10b), which is the first law of thermodynamics for a system: δQ  δW ¼ dE For an ideal gas the only work is due to volume change and thus δW ¼ pd–V . In the intensive form δq ¼ de þ pd– v ¼ dðh  p–vÞ þ pd–v ¼ dh  –vdp By following the change of a gas particle, one essentially is tracking the change of a system. Thus, the time rate change regarding energy can be written using substantial derivatives for the variation and the exact differentials in the above expression and the following is arrived at, with –v ¼ 1=ρ, q_ ¼

Dh 1 Dp  Dt ρ Dt

(8.5l)

which is identical to Eqn. (8.5j).

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8 Equations of Multidimensional Frictionless Flow

8.4 The Entropy Equation The entropy equation for a system undergoing reversible changes is Tds ¼ de þ pd–v

(2.33)

By following similar procedures leading to Eqn. (8.5l), one obtains T

Ds De D–v ¼ þp Dt Dt Dt

(8.6)

For an isentropic flow, the fluid particle experiences no irreversibility or heat transfer (i.e., the adiabatic condition) and receives no work from, and does no work on, the surroundings. Thus according to Eqn. (8.5l),

Ds D p 1 Dp Dh 1 Dp ¼ eþ ¼  ¼ q_ (8.7)  Dt Dt ρ ρ Dt Dt ρ Dt Eqn. (8.7) can easily be shown by substituting Eqn. (8.5l) with q_ ¼ 0 into Eqn. (8.6). Since the substantial derivative tracks changes of a given fluid particle, Eqn. (8.7) indicates that the entropy remains constant along the fluid path. For a steady flow, this translates into constant entropy along a streamline.

8.5 The Substantial (Eulerian) Derivative The physical meaning of the Eulerian operator deserves further explanation. The first term ∂=∂t on the right-hand side of Eqn. (8.1c) denotes the time rate of change of the property (mass, momentum, energy, etc.) due to unsteadiness within ~ ⋅ r term represents the the differential control volume shown in Fig. 8.1. The V change due to the gradient of the property in the direction of the flow while passing through the control volume and is also called the convective derivative, as the inner (or “dot”) product projects the gradient onto the direction of the flow, with the ~ ⋅ r has the unit of 1/time). projected magnitude being the time rate of change (note V One would visualize that for a given gradient in the direction of flow, the larger the velocity, the larger the rate of change. Then, increasing the flow speed by a factor of two implies twice as much power needs to be added to the fluid to maintain the same temperature rise (i.e., same temperature gradient). ~_ represents the force experienced by ~ =Dt ¼ ρ∂V ~ =∂t þ ρV ~ ⋅ rV Similarly, ρDV the fluid particle due to the unsteadiness within the control volume and the velocity gradient across the control volume (the latter being the convective acceleration). Therefore, even if flow is a steady one, the fluid element accelerates/decelerates (with an increase/decrease in its momentum) while moving from one point to the other (for example, in a converging-diverging nozzle). For flows with negligible friction and body forces, Eqn. (8.2e) indicates that the acceleration of the fluid particle is due to the pressure gradient.

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8.6 Fluid Rotation, Vorticity, and Circulation

217

8.6 Fluid Rotation, Vorticity, and Circulation A special class of fluid motion allows great simplification of governing equations. It is the irrotational flow. To discuss irrotational fluid motion, it is necessary to introduce vorticity. In general, vorticity in a flow arises from rotational motion of the fluid particle. For example, a fluid particle with a rigid-body rotation possesses vorticity. A velocity or momentum boundary layer near a surface comprises a vorticity field resulting from the velocity gradient due to the effect of fluid viscosity. A flow of “idealized” fluid with no viscosity (i.e., an inviscid fluid) does not generate a boundary layer near surfaces, although it may possess vorticity due to rotation. Free vortical motion of real fluids, of which hurricanes are among the examples, results in no vorticity. Therefore, there is no definite relationship between fluid viscosity and vorticity, as the former is a property of the fluid and the latter a property of the flow. For further discussion on vorticity, it serves to first demonstrate how vorticity can be calculated from the velocity field (a property of the flow, not the fluid). To understand and characterize rotational fluid motion, consider a differential fluid element OACB shown in Fig. 8.2, where the velocity at point O is ð~ıu þ~jvÞ and the fluid element at the instant rotates about the z-axis. By following the fluid element, one can describe its rotational and deformational motion as if the fluid element is “pinned” at point O. Also shown in Fig. 8.2 are the line segments (now dashed lines) that have undergone a general motion for over an infinitesimal time interval (dt), with the fluid element now represented by OA’C’B’. The line segments OA and OB have turned through angles dθA counterclockwise and dθB clockwise, respectively, to the locations represented by OA’ and OB’. These angles are related to the velocity of the fluid element as follows: 

 1 ∂v ∂v v þ dx dt  vdt ¼ dt dθA ¼ dx ∂x ∂x

dθB ¼

1 dy



 ∂u ∂u uþ dy dt  udt ¼ dt ∂y ∂y

y C' dθB

B dy

ν

O

Figure 8.2 Schematic showing simultaneous rotation and deformation of fluid element OACD in a two-dimensional xy-plane.

C

B'

dθA

A'

x

u

A dx

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8 Equations of Multidimensional Frictionless Flow

For the differential fluid element, the fluid rotational speed about the z-axis (ωz ) is the average of the angular motion of the segments OA and OB. By observing the right-hand rule,

1 dðθA  θB Þ 1 ∂v ∂u ¼  ωz ¼ 2 dt 2 ∂x ∂y Similarly, the angular speeds about the y- and x-axes are, respectively,

1 ∂u ∂v  ωy ¼ 2 ∂z ∂x and

1 ∂w ∂v  ωx ¼ 2 ∂y ∂z Therefore in Cartesian coordinates, fluid rotational velocity is 





 1 ∂w ∂v ∂u ∂v ∂v ∂u ~    þ~j þ~ k ω ¼ ~ı 2 ∂y ∂z ∂z ∂x ∂x ∂y In cylindrical coordinates (r, θ, z), ~ ω is given by





1 ∂Vz ∂Vθ ∂Vr ∂Vz 1 ∂rVθ 1 ∂Vr ~ ~    ω ¼~ er þ~ eθ þk r ∂θ r ∂r r ∂θ ∂z ∂z ∂r

(8.8)

(8.9)

By using the curl operator (r), 1 ~ ~ ω ¼ rV 2

(8.10)

It is customary to define vorticity as twice the fluid rotation velocity, ~ ~ ζ ¼ 2~ ω ¼rV

(8.11)

A flow field with ~ ω ¼ 0 (and thus ~ ζ ¼ 0) is said to be irrotational. The reader can readily show that one-dimensional flows are inherently irrotational. The fluid particle shown in Fig. 8.2 has also experienced an angular deformation rate on the xy-plane (xy ) equal to

1 dðθA þ θB Þ 1 ∂v ∂u ¼ þ xy ≡ (8.12a) 2 dt 2 ∂x ∂y Similarly, the angular deformation rates on the yz- and zx-planes are, respectively,

1 ∂w ∂v þ yz ¼ (8.12b) 2 ∂y ∂z and

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8.6 Fluid Rotation, Vorticity, and Circulation (a)

219

V T=

y dl

∫cV • dl

dy x dx

Figure 8.3 (a) Schematic showing flow circulation, Γ, over an arbitrary closed contour C; (b) a special case of a closed contour – a parallelogram OACB.

C (b)

y

u + ∂u Δy ∂y C

B Δy

v + ∂u Δx ∂x

v v O

A

x

Δx

zx ¼



1 ∂w ∂u þ 2 ∂x ∂z

(8.12c)

If OA and OB rotate through the same angle and in different senses from their original positions (i.e., respectively from the x- and y-axes), then dθA ¼ dθB and the fluid particle would experience only angular deformation (besides convection, which is the translational motion of point O in Fig. 8.2). In this case ∂v=∂x ¼ ∂u=∂y, ∂u=∂z ¼ ∂w=∂x, and ∂w=∂y ¼ ∂v=∂z; similarly xy ¼ ∂u=∂y, yz ¼ ∂v=∂z, and zx ¼ ∂w=∂x. In general, dθA ≠ dθB and the fluid motion results in a combination of rotation and deformation, that is, ~ ω ≠ 0, xy ≠ 0, yz ≠ 0, and zx ≠ 0. A flow property closely related to rotation is circulation, Γ, defined as ~ ⋅ d~ Γ ≡ ∮ CV l

(8.13)

where the integration is carried out over a closed curve C and d~ l is the differential line segment along C pointing in the counterclockwise direction, as shown in Fig. 8.3a. Consider a special closed curve that is a rectangle lying within C with its detailed velocity components and geometric characteristics shown in Fig. 8.3b. Let ~ ¼~ıu þ~jv at point O. Then the circulation over this small enclosed the velocity be V “rectangular curve” is ΔΓ given by

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220

8 Equations of Multidimensional Frictionless Flow







∂v ∂u ∂v ∂u dy dx  vdy ¼  ΔΓ ¼ udx þ v þ dx Δy  u þ dxdy ¼ 2ωz dxdy ∂x ∂y ∂x ∂y (8.14) Summing up the contribution from all elemental areas like that of the closed curve OACB, one obtains ðð ðð ðð ~ ⋅ d~ ζ z dxdy ¼ ζ z dA (8.15a) l¼ 2ωz dxdy ¼ Γ ≡ ∮ CV A

Equation (8.12) is also the Stokes Theorem that relates the line integral to the surface integral and vice versa. For a two-dimensional flow in the xy-plane, Eqn. (8.14) can be rewritten as ζ z ¼ limdA!0

dΓ dA

(8.15b)

Find the vorticity distribution of an incompressible Couette flow between two infinite flat plates with the velocity distribution given as

EXAMPLE 8.1


  2 y 2 1 ~ ¼~ıu ¼~ı h dp  V h 4 2μ dx

where h is the gap between the two plates with y ¼ 0 at the mid-plane, μ is the fluid viscosity, and dp=dx is the pressure gradient that drives the flow. Solution – For this velocity distribution there is only velocity in the x-direction; then Eqns. (8.8) and (8.11) lead to





∂u ∂v ∂v ∂u ∂u ~ ∂u y dp ~   k ¼ ~ k þ~ k ¼ ~j ζ ¼ ~j ∂z ∂x ∂x ∂y ∂z ∂y μ dx It can be seen that ~ ζ ¼ 0 at the mid-plane, as expected due to the symmetry of the velocity distribution while it reaches the maximum at y ¼  h=2. Comments – The fully developed velocity distribution between two parallel, infinite flat plates arises from the merge of the viscous (velocity/momentum) boundary layers over the plates. Once the two boundary layers merge after some entrance length, the velocity distribution is no longer a function of x because of the requirement for the plates to be infinite. Because the magnitude of the velocity everywhere except at the surface increases with the pressure gradient, one expects vorticity to likewise increase due to an increase in ∂u=∂y (which in this case increases linearly with dp=dx). This example illustrates that viscosity □ causes vorticity and that a parallel flow is not necessarily irrotational.

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8.7 Crocco’s Theorem and Shock-Induced Vorticity

221

Find (1) the vorticity field, and (2) circulation over an arbitrary closed circular curve C with a radius R and centered at r ¼ 0 for the flow field: ~ ¼~ V e θ ðC1 rn Þ.

EXAMPLE 8.2

Solution – Eqn. (8.9) yields



  1 ∂rVθ C1 ∂rnþ1 ~ ω ¼~ k ¼~ k ¼~ k ðn þ 1ÞC1 rn1 r ∂r r ∂r ~ ⋅ d~ Γ ≡ ∮ CV l¼

ð 2π 0

Vθ ðrdθÞ ¼

ð 2π

ðC1 rn ÞðrdθÞ ¼ 2πrnþ1 C1

0

A few special values of n might be of interest. ω ¼ 0 (except at r ¼ 0, which is a singular point) For n ¼ 1, Vθ ¼ C1 =r and ~ and the flow field is irrotational and Γ ¼ 2πC1 ¼ constant everywhere, independent of the radius of the circular curve C. ω ¼~ k ð2C1 Þ, and For the rigid-body fluid rotation, n ¼ 1, Vθ ¼ C1 r, ~ Γ ¼ 2πr2 C1 ¼ 2C1 A. The circulation increases linearly with the area of the enclosing circle. ~ ¼~ For n ≠  1, V e θ ðC1 rn Þ and the flow is rotational and the circulation varies with the diameter of the enclosing circle. Comments – For the steady flow of this example,


 ∂ 1 ∂Vr ~ ∂ ~ eθ þk r⋅V ¼ ~ e r þ~ ⋅ ~ e θ ðC 1 r n Þ ¼ 0 ∂r r ∂θ ∂z Thus the flow is incompressible regardless of the value of n. The rigid-body rotation (n ¼ 1) is a “forced” vortex, while the rotational motion with n ¼ 1 is sometimes called a “free vortex.” The former is typical near the eye of a tornado and the latter is found further away from the eye. A free vortex is in fact free of vorticity. A circular fluid motion does not necessarily make the flow rotational. θ Since the only velocity gradient is 1r ∂rV ∂r , for n ¼ 1 this is zero and thus there is no shear stress throughout the flow field. In this case no assumption needs to be made as to whether or not the fluid is inviscid; the effect of fluid viscosity can simply be neglected. This example also demonstrates that rotational motion and □ viscosity do not necessarily generate vorticity.

8.7 Crocco’s Theorem and Shock-Induced Vorticity Examples 8.1 and 8.2 demonstrate how vorticity can be generated and calculated in incompressible flows. For compressible flows, as in incompressible flows, vorticity

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8 Equations of Multidimensional Frictionless Flow

can be calculated for the given, or known, velocity field. As these examples suggest, vorticity could arise from non-uniform velocity fields, Because of the varying shock strength of a bow shock across which non-uniform velocity changes occur, leading to velocity gradients and thus vorticity. Therefore, vorticity generation occurs due to a curved shock wave even if the upstream supersonic flow is uniform in velocity and thermodynamic properties. Can non-uniformity in other flow properties, such as the entropy field across a detached shock wave that is not uniform along the wave, be related to vorticity? A relationship between the entropy change and vorticity generation is useful in allowing calculating vorticity from a state variable, entropy, rather than a vector field of velocity. To focus on the effects of curvature and a nonuniform entropy field on vorticity generation, neglect the force term (~ f ) in Eqn. (8.2d) so that ~ ∂V ~ ⋅ rÞV ~ ¼  1 rp þ ðV ∂t ρ

(8.16)

Replacing the differentials in the combined first and second laws of thermodynamics, Eqn. (2.33), with the gradient operator yields 1 Trs ¼ rh  – vrp ¼ rh  rp ρ which resembles Eqn. (8.6). Combining this expression with Eqn. (8.16) results in Trs ¼ rh þ

~ ∂V ~ ⋅ rÞV ~ þ ðV ∂t

(8.17)

Incorporating the relationship ht ¼ h þ V 2 =2 and the vector identity 2

  V ~ ~ ~ ⋅r V ~ V  rV ¼r  V 2 into Eqn. (8.17) yields
 ~ ~  rV ~ þ ∂V Trs ¼ rht  V ∂t

(8.18)

Equation (8.18) was due to L. Crocco and is called Crocco’s theorem (Shapiro, 1953; Liepmann and Roshko, 1957). For steady flows,
 ~  rV ~ Trs ¼ rht  V (8.19) Examination of Eqn. (8.19) leads to several physical interpretations.

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8.9 Governing Equations in Terms of Velocity Potential

223

~ ≠ 0, r  V ~ ¼~ (1) In general Trs  rht ≠ 0, and because V ζ ≠ 0 and the flow is rotational. (2) In a uniform flow field (such as the one upstream of the shock system) because ~ ≠ 0, ðr  V ~Þ ¼ ~ ζ ¼ 0. A uniform flow is thus irrotational. rs ¼ rht ¼ 0 and V (3) Thirdly and very importantly, rs ≠ 0 downstream of the shock wave due to its curvature (i.e., non-uniform shock strength) and rht ¼ 0 due to adiabatic conditions even with shock wave curvature (see Chapter 4). From Crocco’s theo~ ¼~ rem, r  V ζ ≠ 0; that is, the flow field behind a curved shock is rotational, even if the upstream flow field is uniform and irrotational. The rotational motion and vorticity is therefore induced by the curved shock. The entropy layer is also the layer within which velocity gradients exist; this conclusion is reached even if the fluid is assumed to be inviscid.

8.8 Velocity Potential and Potential Flow In an irrotational flow ωx ¼ ωy ¼ ωz ¼ 0 and ∂v ∂u ∂u ∂w ∂w ∂v  ¼  ¼  ¼0 ∂x ∂y ∂z ∂x ∂y ∂z

(8.20)

The flow can be described by defining a scalar function  ¼ ð x; y; zÞ such that u¼

∂ ∂x

v¼

∂ ∂y

and w ¼

∂ ∂z

(8.21a)

or alternatively, ~ ¼ r V

(8.21b)

Substituting Eqn. (8.21a) into Eqn. (8.20) yields ∂2  ∂2   ≡0 ∂x∂y ∂y∂x

∂2  ∂2   ≡0 ∂z∂x ∂x∂z

and

∂2  ∂2   ≡0 ∂y∂z ∂z∂y

which satisfies the irrotationality condition. The scalar function ϕ is called the velocity potential and, for this reason, irrotational flow is also called potential flow.

8.9 Governing Equations in Terms of Velocity Potential The governing equations for irrotational flows can be summarized as
 Dρ ~ ¼ ∂ρ þ r ⋅ ρV ~ ¼0 þ ρr ⋅ V Dt ∂t

(8.1d)

 ~ ~
DV ∂V ~ ⋅r V ~ ¼ rp ¼ρ þ ρV Dt ∂t

(8.2e)

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224

8 Equations of Multidimensional Frictionless Flow

rht ¼ 0

(8.22)

For a steady irrotational flow, the momentum equation can be derived in terms of velocity potential and becomes
 ~ ⋅r V ~ ¼ rp ρV (8.23) Recalling the speed of sound defined as a2 ≡

∂p ∂ρ

(3.6) s

Therefore, dp ¼ a2 dρ

or

rp ¼ a2 rρ

(8.23)

and


 2 ~ ⋅r V ~ ¼  a rρ V ρ

(8.24)

The steady-state continuity, Eqn. (8.1d), becomes ~ þV ~ ⋅ rρ ¼ 0 ρr ⋅ V Substituting this relationship for rρ=ρ in Eqn. (8.24) yields
 ~ V ~⋅ V ~ ⋅r V ~¼0 a2 r ⋅ V

(8.25)

In the above derivations, the vector identity 2

  V ~ ~ ~ ⋅r V ~ V  rV ¼r  V 2
 ~ ¼ 0 and And because the flow is irrotational, r  V 2


 V ~ ~ V ⋅r V ¼ r 2 Substituting this expression and Eqn. (8.21a) into Eqn. (8.25) produces





∂ ~∂ ~ ∂ ∂ ~∂ ~ ∂ ∂ ∂ ∂ ∂ ∂ ∂ þj þk þj þk þ þ a2 r ⋅ ~ı  ~ı ⋅ ∂x ∂y ∂z ∂x ∂y ∂z ∂x ∂x ∂y ∂y ∂z ∂z

∂ ~∂ ~ ∂ ~ı þj þk ¼0 (8.26) ∂x ∂y ∂z The first term on the left-hand side of Eqn. (8.26) can also be written as a2 r2 , where   r2 is the Laplacian operator and r2 ¼ ∂2 =∂x2 þ ∂2 =∂y2 þ ∂2 =∂z2 in Cartesian coordinates.

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8.10 Flow with Small Velocity Perturbations

225

8.10 Flow with Small Velocity Perturbations; Linearized Theory Flows over a thin airfoil (illustrated in Chapter 4) or a slender body constitute examples of a class of flows called linearized flow, because, they are physically only slightly perturbed from the uniform flow, and the nonlinear terms involving products of derivatives of ϕ in Eqn. (8.26) are negligibly small compared with linear terms. Consider a slight perturbation to a uniform flow having only velocity component U1 parallel to the x-axis, assuming the perturbation is so small that it can be considered as an isentropic process. Thus, due to the isentropic and irrotational nature of the perturbation, the velocity field can be written using the new velocity potential ϕ by substituting  !  þ U1 x

(8.27)



∂ ∂ ~ ∂ ~ þk þ~j V ¼~ı U1 þ ∂x ∂y ∂z

(8.28)

so that

In Eqn. (8.28), ϕ is now the perturbation velocity potential. The velocities due to perturbation can be written as~ıu þ~jv þ ~ kw such that u0 ¼

∂ 0 ∂ ∂ ; v ¼ ; w0 ¼ ∂x ∂y ∂z

(8.29)

For small perturbations, ∂ ∂ ∂ ≪ U1 0 ; ¼ ≪ U1 ; and w0 ¼ ≪ U1 u ¼ ∂x ∂y ∂z 0 2 w and ≪ ≪1 U1 0

or



u0 2 v0 2 ; ; U1 U1

Writing Eqn. (8.26) in terms of velocity,

 a2 r ⋅ ~ı ðU1 þ u0 Þ þ~jv0 þ ~ kw0  ~ı ðU1 þ u0 Þ þ~jv0 þ ~ kw0 ⋅


 0 ∂ 0 ∂ 0 ∂ ~ı ðU1 þ u0 Þ þ~jv0 þ ~ þw ðU 1 þ u Þ þ v kw0 ¼ 0 ∂x ∂y ∂z

(8.30)

(8.31)

Expanding Eqn. (8.29) leads to 0

  ∂u ∂v0 ∂w0 ∂u0 ∂u0 ∂u0 a2 þ þ þ ðU1 þ u0 Þv0 þ ðU1 þ u0 Þw0 ¼ ðU1 þ u0 Þ2 ∂x ∂y ∂x ∂x ∂y ∂z     0 0 ∂v0 ∂v0 ∂w0 ∂w0 2 ∂v 2 ∂w þ v0 þ v0 w0 þ w 0 v0 þ w0 þ v0 ðU1 þ u0 Þ þ w0 ðU1 þ u0 Þ ∂x ∂y ∂z ∂x ∂y ∂z

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226

8 Equations of Multidimensional Frictionless Flow

0

0 0 ∂u0 ∂u ∂v0 2 ∂v 2 ∂w þ v0 þ w0 þ ðU1 þ u0 Þv0 þ ∂x ∂y ∂z ∂y ∂x 0



∂w ∂v0 ∂w0 ∂u0 þ þ þ w0 v0 þ w0 ðU1 þ u0 Þ ∂y ∂z ∂x ∂z

¼ ðU1 þ u0 Þ

2

(8.32)

By neglecting terms containing products of perturbation velocities, one obtains 0

0

0

 2  0 ∂v0 ∂w0 ∂v0 ∂u0 2 ∂u 0 ∂u 0 ∂u 0 ∂w þ þ þ U1 v þ þ a ¼ U1 þ 2U1 u þ U1 w ∂x ∂y ∂z ∂x ∂y ∂x ∂x ∂z (8.33) It is desirable to use the free-stream value of speed of sound, a1 , to pair with the free-stream velocity U1 , so that Eqn. (8.33) would only consist of the free-stream and perturbation velocities. Recall Eqn. (8.20), that is, the stagnation enthalpy of the irrotational flow is constant throughout the entire flow field. Thus ht ¼ h þ

i 1h 2 2 2 ðU1 þ u0 Þ þ v0 þ w0 ¼ constant 2

(8.34)

However, h ¼ cv T þ p–v ¼ cv–T þ RT ¼ γcv–T ¼

cv– 2 a2 a2 a ¼ ¼ R R=cv– γ  1

(8.35)

Similarly, h1 ¼

a21 γ1

(8.36)

Combining Eqns. (8.34) – (8.36) yields i 1h a2 U2 a2 2 2 2 ¼ 1þ 1 ðU1 þ u0 Þ þ v0 þ w0 þ 2 γ1 2 γ1

(8.37a)

By again neglecting the second-order terms and rearranging, Eqn. (8.37a) becomes a2 ¼ a21  ðγ  1ÞU1 u0

(8.37b)

With this, Eqn. (8.33) can be written as 

a21



U12

0

0 0 ∂v0 ∂w0 2 ∂w 0 ∂u 0 ∂v þ þ a1 ¼ ðγ þ 1ÞU1 u þ ðγ  1ÞU1 u þ ∂x ∂y ∂z ∂x ∂y ∂z 0

0

0 0 ∂u ∂v ∂u ∂w þ þ þU1 v0 þ U1 w0 (8.38a) ∂y ∂z ∂z ∂x

 ∂u0

a21

Dividing through by a21 and noting M1 ≡ U1 =a1 , this equation becomes

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8.11 Boundary Conditions for the Linearized Theory



1  M12

227



 ∂u0 ∂v0 ∂w0 u0 ∂u0 u0 ∂v0 ∂w0 þ þ ¼ M12 ðγ þ 1Þ þ M12 ðγ  1Þ þ ∂x ∂y ∂z U1 ∂x U1 ∂y ∂z



v0 ∂u0 ∂v0 w0 ∂u0 ∂w0 þ M12 þ þ þ M12 (8.38b) U1 ∂y ∂x U1 ∂z ∂x

Further simplification of Eqn. (8.38b), considering that ðu0 =U1 Þ; ðv0 =U1 Þ; and ðw0 =U1 Þ ≪ 1, leads to 

1  M12

 ∂u0 ∂v0 ∂w0 þ þ ¼0 ∂x ∂y ∂z

(8.39)

Equation (8.39) is linear, providing the linearized theory for flows is subject to small perturbations. Judicial decisions need to be made when comparing magnitudes of various terms in Eqn. (8.38b). For example, in transonic flow ðM1 ! 1Þ, M12 ðγ þ 1Þðu0 =U1 Þ is of the order of u0 =U1 ; however, it may not be necessarily smaller   than 1  M12 that approaches zero. Therefore, the ∂u0 =∂x on the right-had side of Eqn. (8.38b) cannot be neglected and for small perturbation, the following equation is applicable for subsonic, transonic, and supersonic flows: 

1  M12

 ∂u0 ∂v0 ∂w0 u0 ∂u0 þ þ ¼ M12 ðγ þ 1Þ ∂x ∂y ∂z U1 ∂x

(8.40)

Substituting Eqn. (8.29) into Eqn. (8.40) yields 

1  M12

 ∂2  ∂2  ∂2  M12 ðγ þ 1Þ ∂ ∂2  þ þ ¼ ∂x2 ∂y2 ∂z2 ∂x ∂x2 U1

(8.41)

Thus for transonic flow, Eqn. (8.41) remains nonlinear. For supersonic and subsonic flows, 

1  M12

 ∂2  ∂x2

þ

∂2  ∂2  þ ¼0 ∂y2 ∂z2

(8.42)

which is linear.

8.11 Boundary Conditions for the Linearized Theory For flow over a body, two types of boundary conditions are necessary. As in boundary layer flows, both surface conditions and conditions far away from the surface are necessary. Consider that the small perturbation is due to a solid surface defined by the following function in Cartesian coordinates: gð x; y; zÞ ¼ 0

(8.43)

on which the normal unit vector is

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228

8 Equations of Multidimensional Frictionless Flow

~ n¼

rg jrgj

(8.44)

~ ⋅~ This is because the flow is tangent to the surface, that is, V n ¼ 0. Thus ~ ⋅ rg ¼ 0 V

(8.45)

~ ¼~ı ðU1 þ u0 Þ þ~jv0 þ ~ kw0 ; therefore Over the surface V ð U 1 þ u0 Þ

∂g ∂g ∂g þ v0 þ w0 ¼0 ∂x ∂y ∂z

(8.46)

For two dimensional flows, the tangency requires that v0 v0 dy ∂g=∂x ¼ ≈ ¼ ∂g=∂y U1 þ u0 U1 dx Because of the small perturbation (i.e., y ≈ 0 on the surface),

dy v0 ð x; yÞ ≈ v0 ð x; 0Þ ¼ U1 dx s

(8.47)

(8.48)

where the subscript s denotes that the derivative is evaluated at the surface. The other boundary conditions, far away from the surface, are u0 ¼ v0 ¼ w0 ¼ 0 or < ∞ ði:e; finiteÞ for y ! ∞

(8.49)

Problems Problem 8.1 Use Eqn. (8.5l) to show that for isentropic flow Ds ¼0 Dt Problem 8.2 Carry out the detailed derivation of Eqn. (8.26). Problem 8.3 The velocity of a steady incompressible laminar boundary layer over a flat plate can be represented by an approximation to the Blasius solution as: uð yÞ ¼ U∞ sin


πy 2δ

where U∞ is the freestream velocity in the x-direction and δ is the local thickness of the boundary layer. Find the vorticity field within the boundary layer and show that the magnitude of vorticity is proportional to the average velocity gradient across the boundary and that it is zero at the edge of the boundary layer.

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Problems

229

Problem 8.4 Consider a stream tube enclosed by a material curve C which moves and evolves with the fluid motion as shown here. With the help of Eqns. (8.2e) and (8.13), that is, ~ ⋅ d~ l and Γ ≡ ∮ CV

~ DV 1 ¼  rp Dt ρ

respectively, show that for an incompressible flow DΓ ¼0 Dt that is, the circulation (and the irrotationality/rotationality) is conserved and this expression is called Kelvin’s theorem.

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9

Applications of Small Perturbation Theory

The small perturbation, or linearized, theory derived in Chapter 8 finds a number of useful applications. This chapter discusses applications such as flows over very thin airfoils, where the velocity perturbation is caused by geometry of the surface, and acoustic wave propagation. Relationships among perturbation quantities are derived. Perturbations of pressure, temperature, and density are small compared with free-stream values, while velocity perturbation is much smaller than the local acoustic speed. Flows of a perfect gas are assumed with no effects of viscous and body forces and heat transfer.

9.1 Pressure Coefficient In Chapter 4, the pressure coefficient was defined while discussing the shockexpansion theory:

p  p1 p  p1 2 p  p1 2 p Cp ≡ 1 ¼ 1 γp1 2 ¼ ¼ 1 (4.57) 2 γM12 p1 γM12 p1 2 ρ1 V1 2 γRT V1 1

where subscript 1 denotes free stream conditions. For a flow subjected to a small perturbation, such as by a thin airfoil as shown in Fig. 9.1, it is desirable to express the value in terms of the perturbation. Specifically, the term p=p1 needs to be replaced with perturbation quantities. Recall that ht ¼ constant throughout the flow field. Thus  0 2 02 02 U1 þ u þ v þ w U12 ¼hþ ¼ constant h t ¼ h1 þ 2 2

(9.1)

For a small perturbation, the heat capacity cp can be assumed to remain constant. Dividing Eqn. (9.1) through by cp leads to  0 2 02 02 U1 þ u þ v þ w U12  T  T1 ¼ 2cp 2cp Substituting cp ¼ γR=ðγ  1Þ into this equation and dividing through by T1 yields 230 Downloaded from https:/www.cambridge.org/core. Columbia University Libraries, on 24 Jun 2017 at 04:20:22, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/9781316014288.010

9.2 Linearized Two-Dimensional Flow Past a Wavy Wall M1 U1

231

v

Figure 9.1 Schematic of streamline pattern and flow regions of a flow past an airfoil. Uniform Flow Region

Perturbed Flow Region

 T γ1 2
0 2 02 02 1¼ ½U  U þ u v w  1 1 2 T1 2a1

(9.2)

By using the small perturbation conditions given by Eqn. (8.30) and the isentropic relationship between pressure and temperature, Eqn. (4.4),

u0 U1

2 0 2 0 2 v w ; ; ≪1 U1 U1 p ¼ p1



T T1

(8.30)

γ=ðγ1Þ (4.4)

Substituting Eqn. (9.2) into Eqn. (9.4) while preserving the first-order term of u0 =U1 , it can be shown that (this is left as Problem 9.1) Cp ¼ 

2u0 U1

(9.3)

Thus the pressure coefficient depends linearly and only on the x component of the perturbation velocity.

9.2 Linearized Two-Dimensional Flow Past a Wavy Wall It is of interest to find the pressure coefficient on a two-dimensional arbitrarily shaped surface (e.g., of an airfoil) in the xy-plane (as shown in Fig. 9.2) where the surface protrusion is very small compared with the characteristic dimension in the x direction. Because the protrusion (or elevation) of the surface y can be represented by a Fourier series of sinusoidal functions of x, it is instructive to consider a representative function given by y ¼ f ð xÞ ¼  sin κx

(9.4)

where κ ¼ 2π=L is the wave number with  and L (with  ≪ L) denoting, respectively, the amplitude and the wavelength of the representative wavy wall (as shown in Fig. 9.3) and κ is thus the wave number. For two-dimensional subsonic or supersonic flow and M1 not close to unity,

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232

9 Applications of Small Perturbation Theory y

M1 U1

Figure 9.2 A two-dimensional, arbitrarily shaped surface in the xy-plane (an airfoil is shown).

y = yU (x)

x y = yL (x)

y M1 U1

∋ 0. Equation (9.31) is a linear hyperbolic wave equation and its solutions can be represented by either one of the two arbitrary functions, f ð x  λyÞ and gð x þ λyÞ. The general solution is therefore

k = 2π/L; ∋ 0, p0 > 0, and u0 > 0; that is, the fluid passed by the wave moves in the same direction as the wave. Following similar reasoning, a rarefaction (or expansion) wave causes the fluid passed by the wave to move in the opposite direction as f ð x  a1 tÞ < 0. These phenomena have been described in Chapter 3, where infinitesimal movements of

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9.6 Physical Acoustics in One Dimension

251

the piston-tube device generate compression/expansion waves that travel at the acoustic speed and cause the fluid to move in the same/opposite direction. It is of interest to be able to predict the trajectory of the fluid particle affected by the traveling wave. Let x1 and ξ be, respectively, the initial unperturbed position and the displacement of the fluid particle. Since the choice of x1 is arbitrary, the solution for it represents trajectories of any given fluid particle throughout the fluid. Then the particle position is xð t Þ ¼ x1 þ ξ ð x1 ; t Þ

(9.56)

Following Eqn. (9.54c), the perturbed velocity of the particle becomes u0 ¼ a1 f ðx1 þ ξ  a1 tÞ

(9.57)

Therefore, ξ ð x1 ; t Þ ¼ a1

ðt

f ðx1 þ ξ  a1 τÞdτ

0

where τ is the running variable for time. Because the displacement is due to the passing of the wave, dξ ¼ u0 dτ, one can expect ξ ≈ u0 τ ≪ a1 τ. Then the above expression becomes ξ ð x1 ; t Þ ¼ a1

ðt

f ðx1  a1 τ Þdτ

(9.58)

0 0

Similar to H, ρ0 , p0 , and u , ξ ðx1 ; tÞ also satisfies the wave equation. By using a new variable η ¼ ð x  a1 τÞ, Eqn. (9.58) becomes ξ ð x1 ; t Þ ¼ 

ð x1 a1 t

f ðηÞdη

(9.59)

x1

Further manipulation leads to ð x1 a1 t ð x1 ξ ð x1 ; t Þ ¼  f ðηÞdη þ f ðηÞdη 0

(9.60)

0

Since the wave (disturbance) will be on η ¼ x1  a1 τ ¼ 0, f ðηÞ ¼ 0 for η > 0. For the wave propagating in the þx direction, x1 > 0. Therefore, the first integral results in ξ ðx1 ; tÞ ¼ ξ ðx1  a1 tÞ. Figure 9.9 demonstrates the particle displacement (or trajectory) during the passing of a perturbation wave. The following example describes the effect of a sinusoidal wave.

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252

9 Applications of Small Perturbation Theory t x–a1t = – L x–a1t =– 3 L 4

Trajectory of particle initially at x = x1 for t = 0

x–a1t = –1 L 2

characteristics

x–a1t = – 1 L 4 x–a1t = 0 t = x1/a1 x

x1 L x=–L

Figure 9.9 Particle displacement (or trajectory) during the passing of a perturbation (or acoustic) wave.

EXAMPLE 9.5

Find the expression of fluid particle displacement in a long piston-tube device (in the þx direction, with the piston position initially at x ¼ 0). The piston movement causes a density disturbance described by H ¼  sin κð x  a1 tÞ where  ≪ 1 and κ ¼ 2 π=L is the wave number, with L being the period, or wavelength, of the wave. Solutions – Following Eqn. (9.55), 0

u ¼ a1 f ð x  a1 tÞ ¼ a1 H ¼ a1  sin κð x  a1 tÞ For a fluid particle initially located at an arbitrary location x ¼ x1 in the undisturbed medium, the disturbed particle location is ðt ðt ξ ðx1 ; tÞ ¼ a1  sin κðx1  a1 τÞfdτ ¼ a1  sin κðx1  a1 τÞdτ 0

þa1

ð x1 =a1

x1 =a1

 sin κðx1  a1 τ Þdτ

0

where the second integral is equal to 0 because between t ¼ 0 and x1 =a1 no disturbance has arrived at x1 yet. Thus

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Problems

253

   L 2π 1  cos ξ ðx1 ; tÞ ¼ ½cosκðx1  a1 τ Þtx1 =a1 ¼ ð x1  a1 t Þ κ 2π L or ξ ð x1 ; t Þ ¼ 

ð x1 a1 t

ð x1 a1 t

  sin κηdη ¼  ½cosκηx01 a1 t κ   L 2π 1  cos ¼ ð x1  a1 t Þ 2π L

f ðηÞdη ¼ 

0

0

The particle displacement (or trajectory) is plotted in Fig. 9.9.

□

Problems Problem 9.1 Derive Eqn. (9.3); also show as an intermediate step: p u0 ≈ 1  γM12 p1 U1 Problem 9.2 Following the condition given in Example 9.1, find the pressure fluctuation over the surface. Problem 9.3 Show that for the two-dimensional flow, with free-stream velocity U1 parallel to the x-axis, over the sinusoidal surface, y ¼  sin kx would reach sonic for the free-stream Mach number given by M12 ðγ þ 1Þk 2M12 ðγ þ 1Þπk=L ¼ 1 or    3=2 ¼ 1 3=2 1  M12 1  M12 Problem 9.4 Show steps leading to Eqn. (9.22) for small M1 and show that for very weak perturbations, Eqn. (9.22) simplifies to Eqn. (9.16). Problem 9.5 Carry out the derivation leading to Eqn. (9.26). Problem 9.6 An airfoil shown in the figure is placed in a supersonic flow (M1 > 1) at an angle of attack, α ≪ 1 (i.e., 2 t=c ≪ 1), to the free-stream. Find the lift and drag coefficients using the small perturbation theory. y 1 M1 2

3 2α α

x

t 4 c

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254

9 Applications of Small Perturbation Theory

Problem 9.7 A flat-bottomed airfoil depicted in the figure has its upper surface described by the following expression: y

M1 > 1 α

1

x

–1

  yu ¼ b 1  x2  αx for  1 ≤ x ≤ 1 where the effect of the angle of attack α is included (α measured using the flat lower surface, as shown in the figure), α ¼ αo with  ≪ 1, and b is a constant. (a) Use the small disturbance theory to find Cp , u, and v around the airfoil. (b) Sketch lines of u ¼ constant ½2U∞ b=λ showing the values of the constant on lines through the leading edges on both the upper and the lower surfaces, and the line through the midpoint of the airfoil on the upper surface. (c) Indicate whether the waves of the leading edge are compression or expansion waves. Problem 9.8 Show that for acoustic waves u0 ¼ 

p0 ρ1 a1

where the + and the − signs are for the right- (the F family) and left-running (the G family) characteristics, respectively. Problem 9.9 For an acoustic wave propagating in an ideal gas that experiences a small local density disturbance, as expressed by H ¼ ðρ  ρ1 Þ=ρ1 ¼ ρ0 =ρ1 , with jHj ≪ 1, where the subscript 1 denotes the undisturbed state: (a) Find the change in local speed of sound, ða  a1 Þ=a1 , in terms of γ and H. (b) If the pressure is assumed constant, find an expression of ða  a1 Þ=a1 . Assess the validity of the constant-pressure assumption.

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Problems

255

Problem 9.10 Following Problem 9.9, find an expression of ða  a1 Þ=a1 in terms of γ and K, where K ¼ ð p  p1 Þ=p1 ¼ p0 =p1 , with jKj ≪ 1. Figure for Problem 9.11. Problem 9.11 The piston motion in the piston-cylinder device is described by the following equation: L Up

up

¼ a1 sin ωt; 0 ≤ ωt ≤ 2 π ¼ 0; ωt > 2 π

where ω ¼ a1 κ, a1 is the acoustic speed of the undisturbed gas and κ is the spatial frequency or the wave number. (a) Find the velocity distribution in regions I and II. (b) Would it be possible to arrange the conditions such that no wave in region III is reflected from the piston end of the tube? Problem 9.12 Derive Eqn. (9.26).

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10

Method of Characteristics for Two Independent Spatial Variables

Chapter 9 described solving the governing equations for small perturbations, where two types of characteristics were discussed: two-dimensional steady supersonic flow and unsteady one-dimensional flow. It was also shown that flow properties are constant along characteristics (or characteristic lines) and there exist two families of characteristics: the left-running and the right-running. Examples given in Chapter 9 consist of one single family of characteristics, where only simple wave/ flow regions exist. Practical applications, such as flow in two-dimensional channels, often see both the right- and left-running characteristics intersect to form non-simple wave (or non-simple flow) regions. In unsteady one-dimensional flow, where a wall or surface produces reflected waves propagating in opposite direction to the original disturbances, non-simple flow regions form. In these non-simple flow regions, the method of characteristics (MOC) can be used to solve the flow problem. The MOC takes advantage of the constant properties along the characteristics to solve the nonlinear governing equations for which analytic solutions are rare. This chapter focuses on this method applied to two-dimensional flows. Chapter 11 will be devoted to the MOC in one-dimensional unsteady flows.

10.1 Characteristics in Steady Two-Dimensional Irrotational Flow For the irrotational flow field, it is desirable to define a velocity potential Φ such that ∂Φ ~∂Φ ~ ∂Φ ~ ¼ ~iu þ~jv þ ~ þj þk V kw ¼ ~i ∂x ∂y ∂z

(10.1)

Note that the velocity components, u, v, and w are not necessarily small perturbations as those described Chapter 9. The goal is to find a solution for the partial differential equations governing Φ, as in the case for the perturbation velocity potential . Recall the continuity, irrotationailty, and momentum (i.e., Euler) equations:
 ~ ¼0 r ⋅ ρV (8.1e)

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10.1 Steady Two-Dimensional Irrotational Flow

257

~ ¼0 rV

(8.11)

2


 ~ ⋅r V ~ ¼ rp or r V þ 1 rp ¼ 0 ρV ρ 2

(8.22)

In terms of the velocity potential, the continuity and momentum equations become

2

∂ Φ ∂2 Φ ∂2 Φ ∂Φ ∂ρ ∂Φ ∂ρ ∂Φ ∂ρ ρ þ þ þ 2 þ 2 þ ¼0 (10.2) ∂x2 ∂y ∂z ∂x ∂x ∂y ∂y ∂z ∂z "

2 2 # ρ ∂Φ 2 ∂Φ ∂Φ þ þ dp ¼  d 2 ∂x ∂y ∂z

(10.3)

To arrive at an equation for Φ, it is desirable to eliminate ρ from these equations. The irrotational flow is also homentropic (r ⋅ s ¼ 0), for which a2 ¼ ð∂p=∂ρÞs . Then Eqn. (10.3) becomes "

2 2 # ρ ∂Φ 2 ∂Φ ∂Φ dρ ¼  2 d þ þ (10.4) 2a ∂x ∂y ∂z Therefore,   ∂ρ ρ ∂Φ ∂2 Φ ∂Φ ∂2 Φ ∂Φ ∂2 Φ ¼ 2 þ þ ∂x a ∂x ∂x2 ∂y ∂x∂y ∂z ∂x∂z

(10.5a)

  ∂ρ ρ ∂Φ ∂2 Φ ∂Φ ∂2 Φ ∂Φ ∂2 Φ ¼ 2 þ þ ∂y a ∂x ∂y∂x ∂y ∂y2 ∂z ∂z∂y

(10.5b)

  ∂ρ ρ ∂Φ ∂2 Φ ∂Φ ∂2 Φ ∂Φ ∂2 Φ ¼ 2 þ þ ∂z a ∂x ∂z∂x ∂y ∂z∂y ∂z ∂z2

(10.5c)

Substituting these derivatives into Eqn. (10.2) yields " " " # # # 1 ∂Φ 2 ∂2 Φ 1 ∂Φ 2 ∂2 Φ 1 ∂Φ 2 ∂2 Φ 1 2 þ 1 2 þ 1 2 a ∂x ∂x2 a ∂y ∂y2 a ∂z ∂z2



      2 ∂Φ ∂Φ ∂2 Φ 2 ∂Φ ∂Φ ∂2 Φ 2 ∂Φ ∂Φ ∂2 Φ   ¼0 a2 ∂x ∂y ∂x∂y a2 ∂x ∂z ∂x∂z a2 ∂y ∂z ∂y∂z

(10.6)

Equation (10.6) is the velocity potential equation. It is nonlinear, as opposed to the linear small perturbation equations for subsonic and supersonic flows, Eqn. (8.42). It is further desirable to express the speed of sound a in Eqn. (10.6) in terms of the velocity potential. Taking advantage of Eqn. (8.37a) while noting that U1 is now

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258

10 Method for Two Independent Variables

absorbed in u ¼ U1 þ u0 and taking the state 1 to be the reference state, which is also the stagnation state (U1 ¼ 0 in Eqn. (8.37a)). Therefore, Eqn. (8.37) adapted for the velocity potential is "

2 2 # γ  1 ∂Φ 2 ∂Φ ∂Φ 2 2 a ¼ a1  þ þ (10.7) 2 ∂x ∂y ∂z where a1 is the speed of sound of some reference state. The solution procedure for the flow field involves (i) solving Eqns. (10.6) and (10.7) for Φ subject to boundary conditions, (ii) determining the velocity field by ~ ¼ rΦ, (iii) calculating the speed of sound and local Mach using Eqn. (10.1), V number using Eqn. (10.7) and the velocity determined, and finally (iv) calculating all other state variables, p, T, and ρ using the isentropic relationships derived in Chapter 4. For a quick validation of Eqn. (10.7) consider incompressible flows, for which a ≪ u; v, and w and a ¼ a1 . Then Eqn. (10.7) is reduced to the Laplace’s equation: ∂2 Φ ∂2 Φ ∂2 Φ þ 2 þ 2 ¼0 ∂x2 ∂y ∂z For compressible two-dimensional flows, Eqn. (10.6) becomes " " # #   1 ∂Φ 2 ∂2 Φ 1 ∂Φ 2 ∂2 Φ 2 ∂Φ ∂Φ ∂2 Φ 1 2 þ 1 2  ¼0 a ∂x ∂x2 a ∂y ∂y2 a2 ∂x ∂y ∂x∂y

(10.8)

(10.9)

or  1

   u2 ∂u v2 ∂v 2 uv ∂u þ 1   2 ¼0 a ∂y a2 ∂x a2 ∂y

(10.10)

Equations (10.9a) and (10.10) remain nonlinear, and analytical solutions to nonlinear equations in general do not exist. A numerical approach to solving the nonlinear equation is to use the method of characteristics (MOC). The mathematical theory of MOC indicates (i) that two-dimensional steady flows characteristics, or   Mach lines, exist only for M > 1, that is, u2 þ v2 =a2 > 1 for which the equation is hyperbolic (Courant and Friderichs, 1948), (ii) that the normal derivatives of dependent variables on a characteristic may be discontinuous although they must be continuous, and (iii) that the dependent variables must satisfy a relation called the compatibility relation (Zucrow and Hoffman, 1976; Thompson, 1972). So far the concept of characteristics has been encountered while dealing with problems of isentropic turning (Chapter 4) and acoustic propagation (Chapter 9). Systematic approaches indeed leads to the following compatibility relations (Thompson, 1972):  ðMÞ þ δ ¼ Q ¼ constant ðalong right-running or ξ characteristicsÞ

(4.52a)

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10.1 Steady Two-Dimensional Irrotational Flow

 ðMÞ  δ ¼ R ¼ constant ðalong left-running or η characteristicsÞ

259

(4.52b)

where  ðMÞ and δ are, respectively, the local Prandtl-Meyer function/angle and flow angle (or wall angle if bounded) relative to a reference coordinate. These compatibility relations are derived in the following by using a systematic approach, where the streamline coordinates are used. When deriving or using Eqns. (4.52a) and (4.52b), the wall surface is considered to be streamlined. Cautions are needed for the difference between the angle δ used in Chapter 4 and here for the method of characteristics. In Chapter 4, δ is the turning angle of the wall surface relative to the flow. For the method of characteristics, δ is the general flow angle measured from a reference coordinate (e.g., the Cartesian coordinate system) and δ > 0 (or δ < 0) if it is counterclockwise (or clockwise). Therefore, for MOC, if a wall turns away from the flow and in a counterclockwise direction, the flow angle is considered positive. Recall that the Prandtl-Meyer function is ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M2  1 dM γþ1 γ1 1 ¼ tan ðM2  1Þ  tan 1 M2  1  ðM Þ ¼ γ1 2 γ1 γþ1 M M 1þ 2

(4.47f) Because the flow direction in general changes throughout the flow with respect to the reference coordinate, it is more convenient to adopt the natural coordinate system, in

n

y

n

Δs

s Δn ∂Δ n + ∂

δ+

ω+

∂ω Δn ∂n

∂δ Δn ∂n

Δn +

ω

Δn

δ

s

∂Δn Δs ∂s

s

Δs x

Figure 10.1 Schematic showing the flow and the coordinate systems in the xy-plane; the flow direction is along the s-coordinate and is counterclockwise from the x-coordinate resulting in a flow angle δ > 0. The n-coordinate is normal to the flow everywhere.

R

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260

10 Method for Two Independent Variables

which the two coordinates are in the streamline direction and the other one is normal to it, that is, ðs; nÞ, as the velocity component in the normal direction is identically zero as the definition of streamlines requires. The flow and the coordinate systems are depicted in Fig. 10.1, where the flow direction is the s-coordinate and, as depicted, is counterclockwise from the x-coordinate resulting a flow angle δ > 0. The irrotationality condition, Eqn. (8.11), in the natural coordinate system is developed as follows. For the differential control volume shown in Fig. 10.1, the circulation ΔΓ can be calculated by following the procedure leading to Eqn. (8.14). Let ω be the velocity component along the streamline. Thus



∂ω ∂s Δn Δs þ Δn (10.11) ΔΓ ¼ þωΔs  ω þ ∂n ∂n Geometric consideration of Fig. 10.1 leads to ∂ðΔsÞ=∂n ¼ Δs=R, with R ¼ Δs=Δδ being the radius of the streamline curvature. Then ΔΓ becomes ΔΓ ¼ 

∂ω ∂Δs ∂ω ∂Δs ΔnΔs  ω Δn  ðΔnÞ2 ∂n ∂n ∂n ∂n

Neglecting the third-order term, ΔΓ ¼ 





∂ω ∂δ ∂ω ∂δ ∂ω ∂δ ΔnΔs þ ω ΔsΔn ¼  ω ω ΔsΔn ¼  ΔA ∂n ∂s ∂n ∂s ∂n ∂s

Irrotationality (ΔΓ ¼ 0) requires: ∂ω ∂δ ω ¼0 ∂n ∂s

(10.12)

Adapting Eqn. (4.47a) by replacing V in the Cartesian coordinate system with ω in the natural coordinate system, one arrives at pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dω M2  1 ¼ d ω pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi With this expression and tan μ ¼ 1= M2  1, Eqn. (10.12) becomes tan μ

∂ ∂δ  ¼0 ∂n ∂s

(10.13)

(10.14)

By using Fig. 10.1 again, one can rewrite the continuity and momentum equations, Eqn. (8.1e) and (8.22), along the streamline as ρuΔn ¼ constant

(10.15)

∂u 1 ∂p þ ¼0 ∂s ρ ∂s

(10.16)

u

It is left as an exercise (Problem 10.1) to combine Eqn. (10.15) and (10.16) by similarly using the definitions of the Mach wave angle and the Prandtl-Meyer function to show that

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10.1 Steady Two-Dimensional Irrotational Flow

∂ ∂δ  tan μ ¼0 ∂s ∂n

261

(10.17)

The difference and the sum of Eqn. (10.14) from Eqn. (10.17), respectively, are ∂ ∂ ð þ δÞ  tan μ ð þ δÞ ¼ 0 ∂s ∂n

(10.18a)

∂ ∂ ð  δÞ þ tan μ ð  δÞ ¼ 0 ∂s ∂n

(10.18b)

Thus Eqns. (10.18a) and (10.18b) result from combining continuity, momentum conservation along the streamline, and the requirement for irrotationailty in the natural coordinate system. The compatibility relation has yet to be found on the characteristics (or Mach lines). To do so, an attempt is made by examining the possibility of finding the derivatives of ð  δÞ and ð þ δÞ in the direction of the characteristics.

(a)

y η or Left-running characteristic at (δ+μ) to x-axis

n

Streamline μ

s, ν δ μ

ξ or Right-running characteristic at (δ – μ) to x-axis

x (b)

Figure 10.2 (a) The axes ξ and η are at the angles μ and þμ to the streamline, where μ ¼ sin 1 ð1=MÞ and therefore the axes ξ and η are Mach lines; (b) details of the ξ and η coordinates relative to the flow (or s and n) coordinates.

n

η (ν–θ = constant) Δη μ μ Δξ

Δs

Δn s Δn ξ

Δn Δs = tan μ (ν+θ = constant)

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262

10 Method for Two Independent Variables

The coordinate system ðξ; ηÞ relative to the streamline is shown in Fig. 10.2, where δ is the local inclination angle of the flow (or streamline) with respective to a reference coordinate (here the x axis). The axes ξ and η are at the angles μ and þμ to the streamline, where μ ¼ sin 1 ð1=MÞ and therefore the axes ξ and η are Mach lines. The derivatives of an arbitrary scalar function f in the coordinate system ðξ; ηÞ can be found as follows. The change in f from point 1 to point 2 in Fig. 10.2a can be found either simply along the ξ direction or along the natural/streamline coordinates; the two results should be equal. Therefore Δf ¼

∂f ∂f ∂f Δξ ¼ Δs þ Δn ∂ξ ∂s ∂n

Dividing this equation through by Δs leads to ∂f Δξ ∂f ∂f Δn ¼ þ ∂ξ Δs ∂s ∂n Δs From the geometry shown in Fig. 10.2b, this expression becomes sec μ

∂f ∂f ∂f ¼  tan μ ∂ξ ∂s ∂n

(10.19a)

Similarly, the derivative of f in the η direction can be shown to be sec μ

∂f ∂f ∂f ¼ þ tan μ ∂η ∂s ∂n

(10.19b)

Comparing Eqns. (10.18) and (10.19), one concludes that (note that secμ≠0) ∂ ð þ δ Þ ¼ 0 ∂ξ

(10.20a)

∂ ð  δÞ ¼ 0 ∂η

(10.20b)

These results indicate that  þ δ ¼ Q ¼ constant along a ξ characteristic ðright-running Mach lineÞ (10.21a)   δ ¼ R ¼ constant along an η characteristic ðleft-running Mach lineÞ (10.21b) Equations (10.21a) and (10.21b) are the compatibility relations along characteristics. They provide solution algorithms for two-dimensional, steady, isentropic flow fields. They constitute two equations for two dependent variables,  and δ – the PrandtlMeyer angle (a function only of Mach number) and the flow inclination from the reference coordinate (x-axis in the Cartesian coordinate system. Thus the original three differential equations (8.1e, 8.11, and 8.22) for three variables (u, v, and ρ) have been reduced to two algebraic equations with two variables. In Eqn. (10.21a) and

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10.1 Steady Two-Dimensional Irrotational Flow

263

R1 = ν1–δ1 (η characteristic)

(a)

η characteristic Zone of influence of point 1

Initial value line (or data curve)

Zone of influence of point 3 w3 (ν3, δ3)

) ν , δ1 w1 ( 1

μ3 3

A

μ1 a

μ1

Q 1=

μ3

ξ characteristic

ν 1+δ 1 2 –δ 2

1 R

2

=ν

Zone of influence of point 2 w2 (ν2, δ2)

μ2 μ2

Domain of dependence of point 3

2 Q2 (ν2 + δ2) (ξ characteristic) b

B Initial value line (or data curve)

(b)

η characteristic ξ characteristic

A

3 η characteristic

a 1

3′

ξ characteristic

1′ 2′ 2

b B

Figure 10.3 (a) Schematic showing zones (or domains) of dependence and influence of various points in the flow field and characteristics with their Riemann invariants; (b) choosing a smaller zone of influence (region 1′-2′-3′) than region 1-2-3 in Fig. 10.3a improves computational accuracy.

(10.21b),  þ δ and   δ are called Riemann invariants. The results thus obtained,  and δ, can be used to determine the velocity field, with  providing Mach number M (and speed) and δ, the flow direction. Thermodynamic properties as functions of M can then be calculated using isentropic relationships or found in Appendix C.

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264

10 Method for Two Independent Variables

The next section illustrates the numerical procedure of the method of characteristics. It is now informative to observe that Eqns. (10.21a) and (10.21b) are identical to Eqns. (4.52a) and (4.52b), respectively, describing simple wave region due to isentropic turning.

10.2 Method of Characteristics The procedure of the method of characteristics is illustrated as follows. Suppose that the values of  and δ are already known along the curve AB shown in Fig. 10.3a, and it is desirable to find the flow properties at an arbitrary point 3 downstream of the curve. The curve AB is also called the data curve or the initial value line. Because of the hyperbolic nature of the problem, it is possible to determine flow properties using only the information originating from some upstream region. The characteristics intersecting at point 3 can be traced back to the data curve such that they are, respectively, the ξ and η characteristics coming from points 1 and 2. Because these characteristics are Mach waves, point 3 lies at the edge of the zones of influence (or zones of action, as described in Section 3.2) of points 1 and 2. Therefore point 3 is also under influence of all points between 1 and 2 on the data line, but not under the influence of those outside of the segment connecting 1 and 2, such as points a and b, as shown in Fig. 10.3a. The domain represented by the triangular region of 1-2-3 is called the domain of dependence for point 3. Similarly, the region downstream of point 3 and bounded by the ξ and η characteristics emanating from it is its zone of influence. One can easily see that the zone of influence of point 3 is also under the influence of points 1 and 2, as the point 3 is the intersection of the characteristics from those two points. Extension of such reasoning to the whole flow field can then be made. Assuming that points 1 and 2 in Fig. 10.3a have the following Mach numbers and flow angle (the latter being counterclockwise from the reference x-axis):

EXAMPLE 10.1

at point 1 – M1 ¼ 2:0 and δ1 ¼ 10° at point 2 – M2 ¼ 1:96 and δ2 ¼ 9° find the Mach number and flow angle at point 3. Solutions – Point 3 is at the intersection of the ξ and η characteristics from points 1 and 2, respectively. Therefore, at point 3:  3 þ δ3 ¼  1 þ δ1 and  3  δ3 ¼  2  δ2 1 1  3 ¼ ð 1 þ  2 Þ þ ðδ1  δ2 Þ 2 2

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10.2 Method of Characteristics

265

1 1 δ3 ¼ ð 1   2 Þ þ ðδ1 þ δ2 Þ 2 2 From Appendix C,  1 ¼ 26:380° and  2 ¼ 25:271° . Therefore,  3 ¼ 25:271° and M2 ≈ 1:98 and δ3 ≈ 9:555° As shown in Example 10.1, the values of variables at point 3 fall between those at points 1 and 2; however, they are not exactly equal to the arithmetic averages of the upstream values. If the coordinates of points 1 and 2 are known, then the location of point 3 can be easily calculated by assuming straight line segments for the characteristics and using the Mach wave angles μ1 and μ2 for slopes of the □ straight lines. Since the location of the characteristics depends on the choice of the downstream location (i.e., the knowledge of the location of characteristics are not known a priori), it is possible to choose point 3’ as shown in Fig. 10.3b. In this case point 3’ is at the edges of the zones of influences of points 1’ and 2’; thus conditions at points 1 and 2 do not affect those at 3’. Such a new choice as shown in Fig. 10.3b is meant to improve accuracy, as the mesh size is decreased. Smaller mesh sizes also make better approximations of the characteristics as straight line segments. Further decreases in the mesh size greatly reduce the error in the prediction of the flow field. A typical characteristic network is shown in Fig. 10.4, bounded by an upper wall and a lower free boundary. The characteristic network thus involves characteristics originating from the wall and from the free boundary. Following Liepmann and Roshko [1957], Fig. 10.5 describes the treatment of all the possibilities. Figure 10.5a shows the scheme for data tabulation for calculating toward point 3. For the interior point (Fig. 10.5b), the data are known and carried by the ξ and η characteristics emanating from points 1 (the constant Q1 ) and 2 (the constant R2 ), respectively. In this case,  3 ¼ ðQ1 þ R2 Þ=2 and δ3 ¼ ðQ1  R2 Þ=2 are to be, and can be, calculated, as demonstrated by Example 10.1. They are in general

η

η

η

s

1

4 μ1 μ1

μ4

5

2

8 μ5 μ5

μ8

μ2 μ2

6

12 9 μ9 μ9 μ6 μ6

7

s

10

η μ10 μ10

s

11

3 ξ

ξ

ξ

s (along free boundary; interface between two fluids or streams)

Figure 10.4 A typical characteristic network bounded by an upper wall and a lower free boundary – interior points: 1, 2, 5, 6, 9, 10; free boundary points: 3, 7, 11; wall points: 4, 8, 12.

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266

10 Method for Two Independent Variables δ

ν Q

R

Q = ν+δ R = ν–δ

(a) Table

1

1

⇒

ν = 1 (Q+R) 2 δ = 1 (Q–R) 2

3

1

3

3

2 δ3 Q1

R2

(b) Interior point

Q1 (c) Solid wall

ν3

Figure 10.5 Known data for calculation toward point 3 – a general quart chart showing possible known data (part a); data known for calculation toward an interior point (part b), a wall point (part c), and a free boundary point (part d). (Source: Liepmann, H. W., and Roshko, A., Elements of Gasdynamics, John Wiley and Sons, New York, 1957)

Q1 (d) Free boundary

1  ¼ ðQ þ RÞ 2

(12.22a)

1 δ ¼ ðQ  RÞ 2

(12.22b)

For point 3 on a wall (Fig. 10.5c), the flow angle δ3 is known, as it is determined by the slope of the solid boundary as well as the constant. The constant Q1 is known (i.e., the data, or the initial value, is supplied by point 1). Then the relationship  3 þ δ3 ¼  1 þ δ1 ¼ Q1 yields  3 , completing the calculation for the flow at point 3. An example of a point at the free boundary is that at the interface of the jet fluid and the ambient fluid outside the supersonic nozzle (such as point 7 in Fig. 10.4). At such points the local Mach number (M3 ) and the Prandtl-Meyer angle ( 3 ) are determined by the isentropic relationship using the ratio of the stagnation to the ambient pressures: pt ¼ p



γ γ  1 2 γ1 M3 1þ 2

Then  3 þ δ3 ¼  1 þ δ1 ¼ Q1 yields δ3 , providing the necessary  3 and δ3 for the flow conditions at point 3.

10.3 Classification of Flow Regions The two-dimensional steady supersonic flow can be categorized into three types. The first type is the general, or non-simple region, where ξ and η characteristics intersect and each of the ξ and η characteristics is curved (see Fig. 10.4 serving as an example). Each of these characteristics corresponds to one value of Q or R (as shown in Fig. 10.6). The values of  and δ at the intersection of the ξ and η characteristics are found by using Eqns. (10.22a) and (10.22b).

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10.3 Classification of Flow Regions

267

R1 R2

η – Characteristics (R = ν – δ)

R3

Q2

Q1

Figure 10.6 Characteristic curves and their associated Riemann invariants in a general, or nonsimple (wave) region.

Q3

ξ – Characteristics (Q = ν + δ)

Q2

Ι

Q1

=

=

η – Characteristics

Q3 R1

R2

R3

uniform flow region

R4

ΙΙ

ΔQ = O Δν = 1 ΔR=–Δδ 2

uniform flow region

Figure 10.7 Characteristic curves and their associated Riemann invariants in a simple (wave) region.

Along a ξ-characteristic, Q = constant, and Eqns. (10.22a) and (10.22b) indicate that the changes in  and δ depend on changes in R following the flow crossing the η characteristics. Therefore, the changes in  and δ are related by the following equation: 1 1 Δ ¼ ðΔQ þ ΔRÞ ¼ ΔR ¼ Δδ 2 2

(10.23a)

Similarly, along an η-characteristic, R = constant, so that 1 1 Δ ¼ ðΔQ þ ΔRÞ ¼ ΔQ ¼ Δδ 2 2

(10.23b)

The second type is the simple region, or simple wave in which either Q or R is constant throughout the entire region. Figure 10.7 illustrates this type of flow for which Q ¼ Qo = constant along all the ξ-characteristics originating from region I and extending into region II. Therefore Q ¼ constant and ΔQ ¼ 0 throughout the entire flow field. The changes in  and δ depend on changes in R following the flow crossing the η-characteristics (i.e., going along the ξ -characteristics). Along an ηcharacteristic, ΔR ¼ 0 and therefore  and δ are constant. As a consequence, both

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268

10 Method for Two Independent Variables

Figure 10.8 Characteristic curves and their associated Riemann invariants in a uniform region. Note Q1 ¼ Q2 ¼ Q3 , R1 ¼ R2 ¼ R3 , and all characteristics are straight lines.

the Mach number and the flow inclination are constant along an η-characteristic. Note that the characteristic is the Mach wave at an angle of sin 1 ð1=MÞ to the flow. Thus in a simple region (simple wave), the η-characteristics (or the ξ-characteristics if R = constant) are straight lines with uniform conditions. It can be seen in Fig. 10.7 that flow in regions I and II are uniform. The third type is the uniform flow region, in which Q = constant and R = constant. Not only along each of the ξ- and η-characteristics but also throughout the entire simple wave region,  and δ are constant. Therefore, all characteristics are straight lines. Figure 10.8 illustrates the simple flow over a flat surface. Consider an air flow at M = 4 turns around a 10° rounded convex corner, as shown in the figure. Find the Mach number and flow direction when the flow has turned through 6° . What is the Mach number when the flow completes the 10° turn. Assume no flow separation due to the turning.

EXAMPLE 10.2

R1 = ν1 – δ = 65.785º (δ = 0º) M1=4.0

R1 R 2 R 3

R4 R 5

R6 (V6– ν1 = 10º) M6=4.88

y x 10º

Solutions – As shown in the figure to the right, Q = constant throughout the corner region. Equation (12.23a) indicates   1  4   1 ¼ Δ ¼ ΔR ¼ Δδ ¼ ðδ4  δ1 Þ ¼  6°  0° ¼ 6° 2 M1 ¼ 4, then  1 ¼ 65:785°  4 ¼ 65:785° þ 6° ¼ 71:785° and M4 ≈ 4:50   Similarly,  6   1 ¼ Δ ¼ 12 ΔR ¼ Δδ ¼ ðδ6  δ1 Þ ¼  10°  0° ¼ 10°  6 ¼ 65:785° þ 10° ¼ 75:785° and M6 ≈ 4:88

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10.4 Characteristics and Approximations by Weak Waves

269

Comments – These results are the same as those of Example 4.13, where Prandtl-Meyer expansion theory was used. □

10.4 Characteristics and Their Approximations by Weak Waves Example 10.2 provides a case where a series of Prandtl-Meyer expansion waves for small turning angles can serve as approximation of the method of characteristics, although characteristics and waves are not the same. That is, a continuous flow over a slowly turning (i.e., curved) wall can be approximated by a series of flat segments, which are successively turned away from the flow through finite, small angles. Over the curved wall (Fig. 10.9a), as many η-characteristics as desired can be drawn for those originating from it. With the flat-segment approximation (Fig. 10.9b), the Prandtl-Meyer expansion fans divide the flow field into regions of uniform flow. Because these regions are simple regions, the changes in the flow (Mach number and direction) are found using Eqn. (10.23), which can be generalized as Δ ¼ Δδ

(10.24)

where the + and − signs are associated for the lower wall (generating ηcharacteristic) and the upper wall (generating ξ-characteristic), respectively. The turning angle δ is positive/negative if it is in the counterclockwise/clockwise from the flow direction. (So in Fig. 10.9 the lower wall has δ < 0 and δ is increasingly negative as the flow turns; therefore Δδ < 0 resulting in Δ > 0 and acceleration/ expansion.) Further approximation involves representing each expansion fan in Fig. 10.9b by a single line, as shown in Fig. 10.9c. Thus the continuous flow change through the expansion fan becomes discontinuous. The angle of the single line is the average of the Mach wave angles of the uniform flow regions upstream divided by it. (Because of the uniform flows, the characteristics in each of the simple regions are parallel straight lines, as shown in Fig. 10.8.) Recall for the shock-expansion theory in Chapter 4, Eqn. (4.47b) can be written as δ þ constant ¼  ðMÞ. The positive/negative sign in Eqn. (4.47a) applies when the flow turning is away from/into the flow direction, for both upper and lower walls. Therefore, if the upper wall turns away from the flow direction, Δ is positive. By following the convention of Eqn. (10.24), an upper wall turning away from the flow direction (generating ξ-characteristic) has a positive value of Δδ as the flow crossing the ξ–characteristics (and along the η–characteristics) would observe   δ ¼ R ¼ constant so that Δ ¼ Δδ > 0, resulting in expansion and acceleration. Similarly, if the upper wall turns into the flow, Eqn. (4.47a) indicate Δ < 0, resulting in flow deceleration and compression. If the lower wall turns into the flow Δδ > 0 and Δ ¼ Δδ < 0, also resulting in flow deceleration and compression. By comparison, Eqn. (10.24) indicates that the lower wall generates η–characteristics, and along ξ characteristics Δ ¼ Δδ < 0 leads to compression and deceleration.

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10 Method for Two Independent Variables

(a) M1, V1

n

η-characteristic

Mn, νn

νn –ν1=Δν=–Δδ Δδ (b) M1, ν1

Mn, νn

(c) M1, ν1

Mn, νn

Figure 10.9 A supersonic flow experiences Prandtl-Meyer expansion: (a) over a slowly turning surface; (b) over a series of flat-segment surfaces, as an approximation of the slowly turning surface; (c) over a further refined surface approximation consisting of a larger number of smaller flat-segment surfaces, where the expansion can be represented by a single line (a Mach wave).

10.5 Application of the Method of Characteristics: Two-Dimensional Supersonic Nozzle Design To achieve supersonic internal flows, such as in the wind tunnel and in the convergingdiverging nozzle, some flow conditions are desirable. In the wind tunnel, uniform,

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10.5 Application of the Method of Characteristics

Initial Expansion section

271

Cancellation section 2, d

10 7 a 1

2

6

Throut area, Ath

b

Cross-sectional area, AD

3 5 4

9 8

11 (also c)

L e

Reflection section Nozzle Length (L)

Figure 10.10 Schematic showing the application of the method of characteristics to solving the flow field within a nozzle, symmetric with respective to the centerline. (The grid numbers are those used in Example 10.3.)

parallel supersonic flows are needed in the test section. The exit flow of a convergingdiverging nozzle for the propulsion purpose is supersonic and parallel to the axis of the nozzle. In both cases, M > 1 and Δδ ¼ 0 at the exit plane of the nozzle. In addition, the wall is designed so that no shock waves exists throughout the nozzle. Therefore isentropic flow assumption is assumed so that the method of characteristics can be used to design the flow field and the shape of the diverging section of the nozzle. Consider the nozzle sketched in Fig. 10.10, which is symmetric with respect to the mid-plane, and where the grid system formed by the intersecting characteristics (or waves) and the nodes is also shown. The accuracy of calculations involved in the calculations increases with the increasing number of characteristics and nodes that are chosen. The following points are noteworthy regarding the constraints given by the design Mach number (MD ) and Δδ ¼ 0 at the nozzle exit plane. (i) An initial expansion section is necessary to accelerate the sonic flow at the throat (located at the dashed line ab in Fig. 10.10) to be supersonic in the diverging section of the nozzle. The initial expansion waves are reflected from the plane of symmetry of the nozzle. (ii) The maximum expansion angle (δmax ) can be arbitrarily chosen; however, along with the number of the waves chosen, the furthest centerline location (point c and thus the segment bc in Fig. 10.10) for wave reflection would then be determined accordingly. The length of the segment bc is the reflection length, Lr . (iii) Because the region downstream of the single-line characteristic cd (the last single-line expansion wave) is a uniform flow region, the flow has the designed Mach number and is uniform and parallel to the plane of symmetry.

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10 Method for Two Independent Variables

(iv) In the cancellation section, the wall is shaped so that the exit flow is parallel to the plane of symmetry. That is, δd ¼ 0. Point d thus determines the length (L) to achieve uniform flow at the nozzle exit (exit plane denoted by the dashed line de) that has MD and is parallel to the plane of symmetry. The length is expected to be dependent upon the number and the lengths of flat segments to achieve the initial expansion. (v) To minimize L, the initial expansion region can be shrunk to a sharp corner. If extension of the nozzle length (such as for the test section of the supersonic wind tunnel) is necessary, the additional wall needs to satisfy Δδ ¼ 0 at the nozzle exit to avoid formation of shock waves. One can imagine that for the upper wall shown in Fig. 10.10, further expansion (Δ ¼ Δδ > 0) results in the exit plane pressure being lower than that of the surroundings, which in turn causes the shock wave. On the other hand, if the nozzle extension is accompanied with Δ < 0, then either compression or shock wave occurs and the flow deflected without achieving the design condition. (vi) The line segment cd has an inclination angle equal to the Prandtl-Meyer angle ( c ¼  d ) for the design Mach number MD (with the “design” Prandtl-Meyer angle  D ¼ sin 1 ð1=MD Þ). Therefore,  c ¼  d ¼  D . For the ξ-characteristics that runs from point 3 to point c,  3 þ δ3 ¼ Q3 ¼ constant ¼  c þ δc However, for the initial expansion (from M ¼ 1), δ3 is also the Prandtl-Meyer angle, so δ3 ¼  3 . Because of the symmetry, δc ¼ 0. Therefore, δ3 ¼

c d D ¼ ¼ 2 2 2

(10.25)

That is, the initial expansion angle has to be half of that of the designed Prandtl-Meyer angle. Furthermore, along the η-characteristics, line cd, the equality  c  δc ¼ Qc ¼  d  δd indicates that δd ¼ δc ¼ 0 and that the wall at point d has to be parallel to the plane of symmetry. (vii) The height of the nozzle exit must satisfy the value of Ae =A corresponding to that for MD . The following example illustrates the procedures that one can follow to design a supersonic nozzle using the method of characteristics with the single-line, finitewave approximation. Design the shape of a two-dimensional supersonic nozzle for air flow to achieve the exit Mach number of 2.0. Use the grid system shown in Fig. 10.10.

EXAMPLE 10.3

Solution – Referring to the numbering system in Fig. 10.10, for MD ¼ 2:0,  c ð¼  11 Þ ¼  d ð¼  12 Þ ¼  D ¼ 26:380° . Therefore, δ3 ¼  D =2 ¼ 13:190° according to Eqn. (12.25). For the initial expansion, three linear segments are chosen so

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10.5 Application of the Method of Characteristics

273

that each of them results in an equal amount of flow deflection, Δδ ¼ δ3 =3 ¼ 4:397° . Therefore, δ1 ¼ 4:397° , δ2 ¼ 8:794° , and δ3 ¼ 13:190° . In addition, δ12 ¼ δd ¼ 0. Following Eqn. (10.21),  þ δ ¼ Q ¼ constant along ξ  characteristics ðline segments 1-4; 2-5-8; and 3-5-9-11Þ   δ ¼ R ¼ constant along η  characteristics ðline segments 4-5-6-7; 8-9-10; and 11-12Þ In the cancellation region, the wall is shaped so that δd ¼ 0. This condition can be achieved by having wall deflection angle of 4:397° each at points 7, 10, and 12. The results using the method of characteristics are tabulated below. Point

þδ¼Q

δ¼R

 ¼ 12 ðQ þ RÞ

δ ¼ 12 ðQ  RÞ

M

μ

1 2 3 4 5 6 7 8 9 10 11 12

8:974° 17:948° 26:380° 8:974° ð¼ Q1 Þ* 17:948° ð¼ Q2 Þ* 26:380° ð¼ Q3 Þ* 26:922° 17:948° ð¼ Q5 Þ* 26:380° ð¼ Q6 Þ* 22:688° 26:380° ð¼ Q9 Þ* 26:380°

0° 0° 0° 8:974° 8:974° ð¼ R4 Þ* 8:974° ð¼ R5 Þ* 8:974° ð¼ R6 Þ* 17:948° 17:948° ð¼ R8 Þ* 17:948° ð¼ R9 Þ* 26:380° 26:380° ð¼ R11 Þ*

4:397° * 8:974° * 13:190° * 8:974° 13:461° 17:677° 17:948° 17:948° 22:164° 22:318° 26:380° 26:380°

4:397° * (wall) 8:974° * (wall) 13:190° *(wall) 0° * (symmetry) 4:487° 8:703° 8:974° ð¼ δ2 Þ* 0° *(symmetry) 4:216° 4:370° ð¼ δ1 Þ* 0° * (design) 0° *(Symmetry)

1.23 1.40 1.54 1.40 1.55 1.70 1.70 1.70 1.85 1.86 2.00 2.00

54:4° 45:6° 40:5° 45:6° 40:2° 36:0° 36:0° 36:0° 32:7° 32:7° 30:0° 30:0°

* Values known prior to calculations for the point.

The height of the nozzle is determined by successively determining the points of intersection between the wall extended from the upstream intersection and the incident wave. For example, point 10 is located where the straight wall extension from point 7 meets the incident wave represented by line 9–10. The extension from point 10 then intersects with wave represented by the straight line 10–11 (or cd) at point 12, thus leading to the height of the nozzle. The final check of the result is done by comparing the isentropic value of Ae =A (= 1.688 for MD ¼ 2:0). The accuracy improves as the number of linear segments □ of the wall is increased (Δδ decreased) in the initial expansion region. The difference between predictions by using the one-dimensional isentropic nozzle flow theory and two-dimensional method of characteristics can be seen from the results of Example 10.3. For instance, the Mach number downstream of the wave cd is uniformly equal to 2.00 (¼ MD ), while its value falls between 1.86 and 2.00. Thus,

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10 Method for Two Independent Variables

on a given cross-sectional plane between points 10 and 12, the Mach number is not uniform, in contrast to the one-dimensional prediction.

Problems Problem 10.1 Derive Eqn. (10.17) using the continuity and momentum equations in the natural coordinate system. (Hint: use the definitions of the Mach wave angle, the Prandtl-Meyer angle/function, and the speed of sound.) Problem 10.2 A supersonic flow of air in a variable channel is shown in the figure. The upper wall of the channel is curved, with the curvature starting at point A and ending at point E, while the lower wall is flat. The lines are the supposed characteristics. The incoming flow (M ¼ 1:5 at the inlet plane AA0 ) is uniform and parallel to the lower wall (and to the upper wall upstream of point A). The inclination angles along the upper wall are δA ¼ 0, δB ¼ 4° , δC ¼ 7° , δD ¼ 10° , and δE ¼ 0, with the negative angles signifying turning away from the flow. Assume the walls are perfectly insulated and frictionless. (a) In which region does the incoming flow remain uniform and parallel? (b) In which region does the Prandtl-Meyer solution hold? (c) Find Mach numbers at points a, b, c, d, and P. What is the Mach number at point E (which will be the Mach number downstream of it)?

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11

Unsteady One-Dimensional Flows and Nonlinear Waves

Previous chapters focused on steady flows, with the exception of linear physical acoustic wave propagation in Chapter 9. This chapter concerns gas dynamic phenomena that are unsteady in nature. To focus on the effect of unsteadiness, flows dependent on only one spatial dimension are considered in this chapter. Three types of flow are of particular interest: moving normal shock and expansion waves, formation of shocks, and interactions between waves. The acoustic wave propagation treated in Section 9.6 is due to infinitesimal disturbances (or amplitudes), and thus is a linear wave. A linear wave is characterized by a constant propagation speed, for which the small perturbation theory adequately describes the phenomenon, and the resultant effect due to interactions between multiple waves is the sum of the individual effects. Waves of finite amplitudes, whether shock or expansion, are nonlinear waves. The resultant effects of nonlinear wave interaction cannot be simply added. With the two independent variables – one spatial and one temporal – the method of characteristics can be used for solving the unsteady wave propagation problem, with the characteristics being in the x-t domain, instead of the x-y domain in the steady two-dimensional flows.

11.1 Introduction Two scenarios of unsteady wave propagation arise: simple and non-simple flow regimes. The first scenario is exemplified by a wave (compression or expansion) moving through a quiescent or uniformly flowing medium contained in a pistoncylinder device with an infinitely long tube, as shown in Fig. 11.1a. In the scenario depicted by Fig. 11.1a, the medium ahead of the shock, propagating with an incident speed (VSI ) is quiescent (Va ¼ 0) while the medium behind the shock travels at a speed Vb . There is only one wave propagating in the medium and no interaction between waves; these types of waves are called simple waves. The flow region due to the passing simple wave is called the simple region (or simple flow regime). An “intuitive” approach is illustrated in the following section for moving normal shocks by considering the relative velocity between the wave and the medium, followed by using the method of characteristics (MOC). The MOC not only solves 275 Downloaded from https:/www.cambridge.org/core. Columbia University Libraries, on 24 Jun 2017 at 04:20:25, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/9781316014288.012

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11 Unsteady One-Dimensional Flows and Nonlinear Waves

(a)

(b)

(a) Va = O

VSI

Vb

(b) p2 < p1

p1

Diaphragm

(c)

Expansion wave

Vexp

Shock wave VSI

p1

(d)

Reflected expansion wave

p2

Figure 11.1 Unsteady one-dimensional wave formation and its propagation – (a) a normal shock wave propagating into a quiescent gas; (b) a diaphragm separating high- and low-pressure gases in a container; (c) after the rupture of the diaphragm, a normal shock wave forms and propagates into the low-pressure gas while an expansion forms and propagates into the highpressure gas; (d) the normal shock and expansion waves have reflected from the end walls of the container and are now propagating into each other.

Reflected shock

(a)

(b) (b)

Vb Mb

x (moving with Vs)

(a)

Vs Ms

pa ρa Ta Va ≠ 0 Ma

(2)

Laboratory observer

W2s = Vs–Vb M2s

(1)

p1 = pa ρ1 = ρa T1 = Ta

W1s = Vs–Va M1s = Ms–Ma

X (Laboratory coordinator)

Figure 11.2 (a) A normal shock wave propagates relative to the laboratory coordinate (or to an observer fixed to the earth, or the X-coordinate); (b) the flow as seen by an observer riding on the shock wave affixed with the x-coordinate.

the unsteady wave propagation problem, but also allows tracking the formation of shock waves. The second scenario results from interaction between two like waves (either both are compression waves or both are expansion waves) or between two unlike waves of moving waves. Wave interaction results in the non-simple flow regime. A typical example can be visualized in a long tube that is divided into high- and lowpressure sections, as separated by a diaphragm (Fig. 11.1b). Upon the rupture of the diaphragm, a compression wave travels into the low-pressure gas while an expansion wave travels into the high-pressure gas (Fig. 11.1c), where the gas ahead of the respective waves possesses the pressure prior to the rupture of diaphragm.

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11.2 A Normal Shock Traveling with a Constant Speed

277

The pressure (and temperature) in the region between the parting waves is to be determined. For simplicity, regions immediately downstream of these two parting waves resemble those left behind by simple waves, but how and where the two regions would interface resulting from waves of two different nature, compression and expansion, will be discussed later in this chapter. Sometime later, the reflected compression and expansion waves travel in opposite directions (Fig. 11.1d) and later pass each other, resulting in a complicated interaction pattern. As a consequence the pressure and temperature in regions behind the reflected waves cannot be readily determined. Tracking the states throughout the tube is tedious and time consuming, especially after multiple reflections have occurred. The method of characteristics is used to describe the flow field due to the non-simple nature of the flow regime.

11.2 A Normal Shock Traveling at a Constant Speed through a Medium In Section 4.3, the phenomenon of a traveling wave was qualitatively described. It was noted that a compression wave “carries along” with it the gas it passes, while the expansion wave “repels” the gas away from it that it passes from the laboratory observer’s point of view. For a quantitative description of these phenomena, first consider a traveling normal shock wave moving with a velocity Vs (the subscript s pertains to the shock here and throughout this chapter) into a gas medium that is flowing at a velocity of Va with respective to the laboratory. The gas behind the shock wave follows the wave with a velocity of Vb with respect to the laboratory, as depicted in Fig. 11.2a, where the coordinate X is fixed to the earth (or the laboratory). For an observer traveling with the same velocity as the shock wave (i.e., “riding” on the wave), the gas is moving into the shock wave with a velocity of W1s ¼ Vs  Va , leaving the gas behind the wave with a velocity of W2s ¼ Vs  Vb , with the subscripts 1 and 2 denoting states ahead of and behind the shock, respectively, as shown in Fig. 11.2b, where the transformed coordinate x is fixed to the shock wave and the observer and the shock wave is stationary; the problem becomes a steady one. Equation (4.11) for the stationary shock wave is applicable and can be rewritten as dp ¼ ρWs dWs

(11.1)

which explains the carrying-along and repelling action of shock and expansion waves, respectively in Chapter 4. In the case of a moving shock, the gas behind the shock is being carried along, with a velocity Vb relative to the X-coordinate. The velocity Vb is called the induced mass motion. In the transformed x-coordinate system, all equations derived in Chapter 4 are applicable, with the requirements: (1) all velocities are relative to the shock wave, and (2) the thermodynamic properties in the gas ahead of the shock are static properties as the shock wave acts to compress the gas it passes. It is important to recall that all static properties are defined as measured by a probe moving at the absolute flow velocity and are independent of the velocity of

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11 Unsteady One-Dimensional Flows and Nonlinear Waves

the observer, whether riding on the wave or standing in the laboratory. Therefore, by referring to Figs. 11.2a and 11.2b, pb p2 Tb T2 ¼ and ¼ pa p1 Ta T1

(11.2)

A special case involves Va ¼ 0 (i.e., quiescent gas ahead of the shock), for which pa ¼ p1 , pb ¼ p2 , Ta ¼ T1 , and Tb ¼ T2 and which automatically satisfies Eqn. (11.2). On the other hand, the stagnation properties are measured by bringing the gas to rest. In the X-coordinate system, the gas at state a is at rest and thus Tta ¼ Ta and pta ¼ pa , while in the x-coordinate system the gas at state 1 needs to be brought to rest for measuring its stagnation properties. Consequently, pt1 > pta and Tt1 > Tta for the simple reason that the gas in state 1 has the relative velocity W1s ¼ ðVs  Va Þ or, alternatively, that the shock (i.e., the moving observer) has a compressive or ramming effect on the gas. As a consequence, W2s ¼ ðVs  Vb Þ < ðVs  Va Þ ¼ W1s and for continuity W2s =W1s ¼ ρ1 =ρ2 ¼ ρa =ρb < 1. One can imagine that if the shock wave has an isentropic ramming effect (which is certainly not true due to the nonisentropic shock compression), an imaginary stagnation pressure exists in the gas ahead of the shock wave, with its value associated with the speed of the moving shock. This imaginary stagnation pressure will decrease across the shock, much like what happens in a gas flow passing through a stationary shock – because the compression due to a shock wave is non-isentropic. The following example illustrates the use of coordinate transformation in solving moving shock problems. A normal shock wave is traveling into quiescent air (with T1 ¼ 216 K and p1 ¼ 19:4 kPa) and causing a pressure ratio of 4.5. What are the air velocity, Mach number, pressure, and stagnation pressure and temperature downstream of the shock as seen by a laboratory observer? Compare the results with those of the stationary shock with the same Mach number and static properties.

EXAMPLE 11.1

Solution – Referring to Fig. 11.2, one finds the same results for this moving shock problem for the following quantities. Note that for quiescent air ahead of the shock, Va ¼ 0, W1s ¼ Vs , and M1s ¼ Ma , where M1s is the air flow Mach number relative to the moving shock. For p2 =p1 ¼ 4:5 Ma ¼ 2:0 For Va ¼ 0, M1s ¼ Ma ¼ 2:0 W1s ¼ Vs ¼ M1s

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi γRTa ¼ 2:0  1:4  287  216 ¼ 590:2 m=s

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11.2 A Normal Shock Traveling with a Constant Speed

279

Therefore, Tt1 =T1 ¼ 1:80 Tt1 ¼ Tt2 ¼ 388:80 K (due to no heat transfer to the one-dimensional steady shock wave) M2s ¼ 0:5774 (this relative to the shock wave) T2 =T1 ¼ 1:6875; T2 ¼ 364:5 K; ρ2 =ρ1 ¼ 2:667; pt2 =pt1 ¼ 0:7209 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi W2s ¼ ðVs  Vb Þ ¼ M2s γRT2 ¼ ð0:5774Þ ð1:4Þð 288 J=kg ⋅ K Þð1:6875  216 K Þ ¼ 221:4 m=s: W2s can also be found by using the continuity equation:



pffiffiffiffiffiffiffiffiffiffiffi p1 ρ T2 W2s ¼ W1s 1 ¼ M1s γRT1 ρ2 p2 T1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  1:6875 ¼ 2  ð1:4Þð 288 J=kg ⋅ K Þð216 KÞ  4:5 W2s ¼ 221:4 m=s

γ  1 2 γ=ðγ1Þ M1 pt1 ¼ p1 1 þ ¼ ð19:4 kPaÞð7:8247Þ ¼ 151:80 kPa 2 (pt1 is the “imaginary” stagnation pressure.) pt2 ¼ 0:7209pt1 ¼ 109:43 kPa (This is the stagnation pressure downstream of a stationary shock with M1s ¼ 2.) Note that ðVs  Vb Þ ¼ 221:4 m=s is the gas velocity relative to the moving shock. It is in the −x direction in the transformed coordinate system. Then pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi Vs ¼ M1s γRT1 ¼ 2:0  ð1:4Þð 288 J=kg ⋅ K Þð216 K Þ ¼ 590:2 m=s 221:4 2s (Check – W W1s ¼ 590:2 ¼ 0:375 ≈

ρ2 ρ1

with ρρ1 ¼ 2:667 given in the above.) 2

Vb ¼ Vs  ðVs  Vb Þ ¼ 590:2 m=s  221:4 m=s ¼ 368:8 m=s Therefore, to the laboratory observer, the gas behind the shock moves in the +X direction, that is, it is being carried along by the wave. The Mach number of the gas flow behind the shock is Mb : pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi Mb ¼ Vb = γRT2 ¼ 368:8 m=s= ð1:4Þð 288 J=kg ⋅ K Þð1:6875  216 K Þ ¼ 0:962 To find the stagnation temperature Ttb one needs to first find Tb . Tb T2 Ta ¼ T1 ; Ta ¼ T1 , and Tb ¼ T2 . Tb ¼ T2 ¼ 364:5 K γ1 Ttb 2 Tb ¼ 1 þ 2 Mb ≈ 1:19 Therefore,

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11 Unsteady One-Dimensional Flows and Nonlinear Waves

Ttb ¼

Ttb Tb ¼ 1:19  364:5 K ¼ 433:76 K Tb

The stagnation pressure behind the traveling shock is then from the table or ¼ pp21 ; pa ¼ p1 , and pb ¼ p2 for Va ¼ 0.

pb pa

pb ¼ 4:5pa ¼ 4:5p1
γ=ðγ1Þ γ1 ptb 2 ≈ 1:81. Therefore, pb ¼¼ 1 þ 2 Mb ptb ¼

ptb pb pa ¼ 1:81  4:5  19:4 kPa ¼ 157:14 kPa pb pa

Note that for the conditions in this example, Mb ≠M2s and ptb > pt2 and Ttb > Tt2 ¼ Tt1 , because the stagnation properties are measured by bringing the gas flow to rest with respective to the observer, and, thus, post-shock stagnation quantities are different, in the X- and x-coordinate systems. One may make a comparison of these results with those of Example 4.3, where a stationary shock faces an incoming flow at M1 ¼ 2:0 with the same T1 and p1 producing pt2 ¼ 109:43 kPa downstream of the shock and □ Tt2 ¼ Tt1 ¼ Tta ¼ 388:8 K. The results of Example 11.1 suggest that the stagnation temperature and pressure downstream for a moving shock are greater than those across a stationary shock with the same strength, β (or same Mach number), as shown by Eqn. (4.18d): β≡

 p2  p1 2γ  2 M1  1 ¼ γþ1 p1

(4.18d)

This comparison of stagnation properties behind the shock is opposite to that ahead of the shock. Why are there such fundamental, qualitative differences in stagnation properties in flows with steady and moving shock waves? To answer this question, it is helpful to consider for examples the one-dimensional nozzle flow with a standing normal shock and the moving shock wave due to the moving piston in a pistoncylinder device. In the case of the steady/standing shock, no work is done on the gas as it passes through the shock; this is inherently implied in the derivations leading to all the shock relations shown in Chapter 4, by the fact that the stagnation enthalpy remains constant across the shock, ht2 ¼ ht1 . In the case of unsteady/moving shocks, a piston moving into the gas is doing work on the gas, resulting in a higher stagnation temperature, and htb > ht1 (and Ttb > Tt1 ), for the same Mach number. Because of the work, the stagnation pressure also increases over that downstream of the steady shock caused by the same Mach number. One can also see that by the coordinate transformation from the X-coordinate to the x-coordinate, energy/work is added/ done to the entire system; that is, moving the originally stagnant air, as in the state a in Fig. 11.2a (i.e., moving the entire reference frame) requires energy. In the pistoncylinder system, the energy/work is added/done by moving the piston. The work

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11.2 A Normal Shock Traveling with a Constant Speed

281

done by the piston in Example 11.1 results in ptb > pt1 , even though the nonisentropic shock process should cause a decrease in stagnation pressure. Before leaving this section, consider a more general situation of Example 11.2 below, where Va ≠ 0 instead of the quiescent air of Example 11.1. Consider all the same conditions as in Example 11.1, except that Va ¼ 100 m=s (in the X-coordinate system, as shown in Fig. 11.2a), that is, the air is moving in the opposite direction of the shock wave.

EXAMPLE 11.2

Solution – For the same pressure ratio p2 =p1 ¼ 4:5 M1s ¼ Ma ¼ 2:0 M2s ¼ 0:5774 pffiffiffiffiffiffiffiffiffiffiffi In the x-coordinate system, W1s ¼ M1s γRT1 ¼ 590:2 m=s ¼ ðVs  Va Þ Vs ¼ 490:2 m=s Vs  Vb ¼ M2s

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi γRT2 ¼ 0:5774  ð1:4Þð 288 J=kg ⋅ K Þð1:6875  216 K Þ ¼ 221:35 m=s Vb ¼ Vs  ðVs  Vb Þ ¼ 268:9 m=s

This Vb ¼ 268:9 m=s is smaller than that in Example 11.1 by exactly 100 m/s, consistent with the fact that the transformed x-coordinate does not have to advance at the same large velocity as in the case where Vb ¼ 0. That is for Va ¼ 100 m=s, all gas velocities is reduced by 100 m/s in the X direction. The static temperature and pressure remain the same as those in Example 11.1. However, the stagnation properties need to be calculated using the new Mach number Mb for the new Vb . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi Mb ¼ Vb = γRT2 ¼ 268:9 m=s= ð1:4Þð 288 J=kg ⋅ K Þð1:6875  216 K Þ ¼ 0:701
γ=ðγ1Þ ptb γ1 2 ≈ 1:388. Therefore, pb ¼¼ 1 þ 2 Mb ptb ¼

ptb pb pa ¼ 1:388  4:5  19:4 kPa ¼ 121:2 kPa pb pa Ttb γ1 2 Mb ≈ 1:098 ¼ 1þ 2 Tb Ttb ¼

Ttb Tb ¼ 1:098  364:5 K ¼ 400:3 K Tb

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11 Unsteady One-Dimensional Flows and Nonlinear Waves

The values of ptb and Ttb are different from those obtained for Va ¼ 0 in Example 11.1. Comments – If Va ¼ 590:2 m=s, then W1s ¼ 0, resulting in a stationary shock. Then the results of Example 4.3 are recovered. In this case there is no need to add energy to move □ the air in the state a in Fig. 11.2a for the shock to be stationary.

11.3 General Relations for a Traveling Normal Shock at a Constant Speed The previous section described an intuitive approach to the traveling normal shock problem by invoking the stationary normal shock relations, with coordinate transformation and proper accounting of relative velocity with respect to both the laboratory and the shock itself. Such an intuitive approach resorts to and reflects understanding of the physical event of a moving shock. Expanding on this understanding, the following describes a formulation of general scenarios of the relative velocity between the shock and the gas ahead of it. Consider the transformed stationary shock problem depicted in Fig. 11.2b. The conservation laws for mass, momentum, and energy are ρ1 ðVs  Va Þ ¼ ρ2 ðVs  Vb Þ

(11.3)

p1 þ ρ1 ðVs  Va Þ2 ¼ p2 þ ρ2 ðVs  Vb Þ2

(11.4)

h1 þ

ðVs  Va Þ2 ðVs  Vb Þ2 ¼ h2 þ 2 2

(11.5)

Equations (11.3) through (11.5) are identical with those for stationary normal shocks given in Chapter 4. Therefore, with similar algebraic manipulations as detailed in Chapter 4, the Rankine-Hugoniot equation for a traveling normal shock can be obtained as 2 2 ðVs  Va Þ2  ðVs  Vb Þ2 ¼ W1s  W2s ¼ ðp2  p1 Þð–v1  –v2 Þ

1 h2  h1 ¼ ðp2  p1 Þð–v1  –v2 Þ 2

(11.6a)

(11.6b)

which is identical with Eqn. (4.71a) for a stationary normal shock. As expected from the reasoning given in Section 4.18, the Rankine-Hugoniot relationship relates thermodynamic quantities that do not depend on whether or not the shock wave is traveling. For a calorically perfect gas, h ¼ cp T and –v ¼ RT=p. With cp ¼ γR=ðγ  1Þ Eqn. (11.6) becomes

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11.3 General Relations for a Traveling Normal Shock

2 3 γþ1 p2 T2 p2 4 γ1 þ p1 5

 ¼ T1 p1 1 þ γþ1 p2 γ1

283

(11.7)

p1

or
ρ2 p2 T1 ¼ ¼ ρ1 p1 T2

γþ1 γ1




γþ1 γ1

p2 p1

þ

þ1

p2 p1

(11.8)

Equations (11.7) and (11.8) are the same as Eqns. (4.72) and (4.73) for a stationary normal shock. For the transformed stationary coordinate shown in Fig. 11.2b, Eqn. (4.18b) is applicable, with the Mach number now being M1s : 2 p2 2γM1s γ1  ¼ γþ1 p1 γþ1

(11.9)

Therefore, sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

γ þ 1 p2 M1s ¼ 1 þ1 2γ p1

(11.10)

and sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

γ þ 1 p2 W1s ¼ Vs  Va ¼ M1s a1 ¼ a1 1 þ1 2γ p1

(11.11)

By using the continuity equation, Eqn. (11.3), one finds W2s ¼ Vs  Vb ¼ and Vb ¼

ρ1 ρ ðVs  Va Þ 1 W1s ρ2 ρ2





ρ ρ ρ 1  1 Vs þ 1 Va also ¼ 1 W1s ρ2 ρ2 ρ2

For the special case where Va ¼ 0, Eqn. (11.11) is reduced to sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

γ þ 1 p2 γþ1 β þ 1 ðfor Va ¼ 0Þ W1s ¼ Vs ¼ a1  1 þ 1 ¼ a1 2γ 2γ p1

(11.12)

(11.13)

(11.14)

Similarly by incorporating Eqn. (4.18e) for β ¼ ðp2 =p1  1Þ, Eqn. (11.13) becomes vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

u



2γ u γþ1 ρ1 a1 p2 2a1 1 t ¼ M1s  1 Vb ¼ 1  Vs ¼ ðfor Va ¼ 0Þ p2 M1s ρ2 γ p1 þ γ1 γ þ 1 p1

γþ1

(11.15)

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11 Unsteady One-Dimensional Flows and Nonlinear Waves

The derivation of Eqn. (11.15) is left as an exercise problem (Problem 11.1). Equation (11.13) indicates that as W1s increases, the gas velocity behind the traveling shock decreases. Such a result helps to explain the results of Examples 11.1 and 11.2. Example 11.2 has a larger W1s and yields a smaller Vb ; it also leads to a smaller Mb , as expected, because the acoustic speed is a function of static temperature, which does not depend on whether or not the shock is traveling. The Mach number of the induced mass action of gas behind the shock observed by the laboratory (X-coordinate) observer is

Vb Vb 2 1 a1 M1s  ¼ ¼ (11.16) Mb ¼ γþ1 M1s a2 ab a2 pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi because a ¼ γRT , a1 =a2 ¼ T1 =T2 and therefore

rffiffiffiffiffiffi 2 1 T1 M1s  (11.17) ðfor Va ¼ 0Þ Mb ¼ γþ1 M1s T2 Attention should be paid to the use of Eqn. (11.13), because values of Vb for Va ¼ 0 and Va ≠ 0 are different for a given ratio, p2 =p1 . Therefore, an expression like Eqn. (11.17) is not readily available for Va ≠ 0, because W1s ≠ Vs and Eqn. (11.15) cannot be used. Example 11.3 illustrates such a difference between values of Vb for Va ¼ 0 and Va ≠ 0. For a very strong normal shock (p2 =p1 ! ∞) traveling into either a quiescent or flowing gas, it can be shown that (left as an end-of-chapter problem) sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 lim Mb ¼ (11.18) γðγ  1Þ p2 =p1 !∞ For comparison, for very a strong stationary normal shock, sffiffiffiffiffiffiffiffiffiffiffi γ1 ðM2 ¼ M2s in Fig: 112bÞ M2 ¼ 2γ

(4.21b)

Once again, as in results in Example 11.1, Mb ≠M2s . For γ ¼ 1:4, Mb ¼ 1:890 and M2 ¼ 0:378. (But as shown in Example 11.2, if Va ¼ W1s then M2s ¼ Mb ¼ 0:378, i.e., the result of a stationary shock with M1 ¼ M1s .) It is interesting to note from these results that the induced gas motion behind the traveling shock can be supersonic for sufficient shock strength! Use Eqns. (11.13) and (11.17) to find out the induced velocities and Mach numbers for the conditions in Examples 11.1 and 11.2.

EXAMPLE 11.3

Solutions – For Example 11.1, Va ¼ 0 and pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a1 ¼ γRT1 ¼ ð1:4Þð 288 J=kg ⋅ KÞð216 K Þ ¼ 295:1 m=s

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11.3 General Relations for a Traveling Normal Shock

Vs ¼ W1s ¼ M1s

Vb ¼

a1 γ



285

pffiffiffiffiffiffiffiffiffiffiffi γRT1 ¼ 590:2 m=s

vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

u 2γ u γþ1 ð1:4Þð 288J=kg ⋅ KÞð216 K Þ p2 t ¼ 1  3:5  0:5 p2 γ1 1:4 p1 p1 þ γþ1

¼ 368:9 m=s vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi

u

u 2γ 2γ u u γþ1 Vb 1 a1 p2 1 T1 p2 γþ1 t ¼ ¼  1 tp  1 Mb ¼ p2 2 γ a2 p1 a2 þ γ1 γ T2 p1 þ γ1 p1

γþ1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
 ffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiv u γþ1 p2 u

u 2γ u γþ1 u 1 þ γ1 p1 1 p2 u

 ¼  1 tp 2 γ1t 2 γ p1 γþ1 p2 p2 p1 þ γþ1 γ1 p1 þ p1

p1

γþ1

where Eqn. (11.7) is used for T1 =T2 or because M1s ¼ Ma ¼ 2, T2 =T1 ¼ 1:6875 from Appendix D can be used directly. Therefore, sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2:8 1 þ 2:4 1 0:4  4:5 2:4  3:5  ¼ 0:962 Mb ¼ 0:4 2 1:4 4:5 þ 2:4 2:4 0:4 ð4:5Þ þ ð4:5Þ Alternatively, Eqn. (11.16) can be used to determine the value of Mb –



rffiffiffiffiffiffi 2 1 a1 2 1 T1 M1s  M1s  ¼ Mb ¼ γþ1 M1s a2 γ þ 1 M1s T2

rffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 1  2 ¼ 0:962 ¼  2:4 2 1:6875 For Example 11.2, the contribution of Va ¼ 100 m=s to Vb is calculated, using Eqn. (X-13). In this case, W1s ¼ ðVs  Va Þ ¼ 590:2 m=s and Vs ¼ 490:2 m=s



ρ ρ p1 T2 ρ Vb ¼ 1  1 Vs þ 1 Va ¼ 1  Vs þ 1 Va ρ2 ρ2 p2 T1 ρ2 Vb ¼

Mb ¼

1:6875 1:6875 ð100 m=sÞ ¼ 268:8 m=s 1  490:2 m=s þ 4:5 4:5

Vb Vb 269:9 m=s 268:8 m=s ¼ ¼ pffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:701 a2 a2 γRT2 ð1:4Þð 288J=kg ⋅ KÞð1:6875  216 K Þ

Comments – The reader can find that for Va ¼ 590:2 m=s, Mb ¼ M2s ¼ 0:577 (stationary □ shock result).

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286

11 Unsteady One-Dimensional Flows and Nonlinear Waves x (b)

Va = 0 Ma = 0 pa ρa Ta

(a)

Vb Mb pb ρb Tb

Vs Ms

(2)

W2s = Vs–Vb p2 ρ2 T2

(1)

V1 = Vs = W1S p1 = pa ρ1 = pa T1 = Ta

X

(b) Vsr Mb pb ρb Tb

(c)

(2) Vc = 0 Mc = 0 pc ρc Tc

W2s = Vsr+Vb p2 ρ2 T2

(3)

V3 = Vsr – Vc = Vsr p3 ρ3 T3

X

Figure 11.3 (a) A normal shock propagating toward a wall, with the flow field seen by an observer riding on the wave (the x-coordinate); (b) the reflected normal shock wave from the wall, with the flow field seen by an observer riding on the wave.

11.4 Reflected Normal Shock Waves The reflected normal shock wave depicted in Fig. 11.1b is now further described in Fig. 11.3 with the details of the shock movements and relevant thermodynamic and flow variables. Fig. 11.3a shows the incident wave, with state ahead of the incident wave having quiescent gas while carrying along with it gas behind it at state b. The reflected wave, shown in Fig. 11.3b, travels into the gas at state b leaving gas behind it at state c. To satisfy the non-moving wall condition, Vc ¼ 0, which then determines the strength of the shock wave, which is such that it brought the gas behind it to rest and is expected to be different from the strength of the incident wave. The following example illustrates the intuitive approach in finding state c, behind the reflected shock by using the stationary shock relations with coordinate transformation. An incident normal shock wave travels in quiescent air (at 100 kPa and 300 K) in a direction perpendicular to a plane wall. The incident wave causes a pressure ratio of 2.60. Find the temperature, pressure, velocity, and Mach number in state c behind the reflected shock wave.

EXAMPLE 11.4

Solution – Referring to Fig. 11.3a, one finds the same results for this moving shock problem for the following quantities. Note that for quiescent air ahead of the shock, Va ¼ 0, W1s ¼ Vs , and M1s ¼ Ma .

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11.4 Reflected Normal Shock Waves

287

For p2 =p1 ¼ 2:6 Msi ¼ M1s ¼ Ma ¼ 1:54 (The subscript si denotes the incident wave.) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi Vsi ¼ W1s ¼ Msi γRT1 ¼ 1:54 ð1:4Þð 288J=kg ⋅ K Þð300 K Þ ¼ 535:6 m=s Therefore, M2 ¼ 0:687 T2 =T1 ¼ 1:347; T2 ¼ 404:1K p2 =p1 ¼ pb =pa ¼ 2:60 as given: W2s ¼ ðVs  Vb Þ ¼ M2

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi γRT2 ¼ ð0:687Þ ð1:4Þð 288J=kg ⋅ K Þð404:1 K Þ ¼ 277:3 m=s: Vb ¼ 258:3 m=s

For the reflected shock, the wall condition requires that Vc ¼ 0. Therefore, V2 ¼ Vsr þ Vb (The subscript sr denotes the reflected wave.) V3 ¼ Vsr The continuity equation, Eqn. (11.3), for the reflected wave becomes ρ2 ðVsr þ Vb Þ ¼ ρ3 Vsr This expression is rearranged to give ρ3 Vsr þ Vb ¼ ρ2 Vsr Note that Vsr þ Vb Msr ¼ pffiffiffiffiffiffiffiffiffiffiffi γRT2 Equation (4.18d) is used to relate ρ3 =ρ2 to Msr : 2

þVb Þ 2 ðγ þ 1Þ ðVsrγRT ρ3 ðγ þ 1ÞMsr Vsr þ Vb 2 ¼ ¼ ¼ 2 þ2 ðVsr þVb Þ2 ρ2 ðγ  1ÞMsr Vsr ðγ  1Þ þ2 γRT2

Rearranging the above expression yields

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288

11 Unsteady One-Dimensional Flows and Nonlinear Waves

ðγ þ 1ÞðVsr þ Vb Þ2 Vsr ¼ ðγ  1ÞðVsr þ Vb Þ2 Vsr þ ðγ  1ÞðVsr þ Vb Þ2 Vb þ 2ðγRT2 ÞVsr þ 2ðγRT2 ÞVb With the values of Vb and T2 , the above equation becomes 2Vsr3 þ 930Vsr2  245; 713Vsr  91; 042; 585 ¼ 0 Vsr ¼ 329:0 m=s Vsr þ Vb 329:0 m=s þ 258:3 m=s M2 ¼ Msr ¼ pffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ≈ 1:455 and M3 ≈ 0:72 γRT2 ð1:4Þð 288 J=kg ⋅ K Þð404:1 K Þ It is noted that ðVsr þ Vb Þ ¼ V2 and pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi V2 ¼ M2 γRT2 ¼ 1:455  1:4  287  404 ≈ 587:0 m=s same as ðVsr þ Vb Þ. p3 =p2 ¼ pc =pb ≈ 2:30; p3 ¼ pc ¼ 2:30  2:6  100 kPa ¼ 598:0 kPa T3 =T2 ¼ Tc =Tb ≈ 1:29; T3 ¼ Tc ¼ 1:29  1:347  300 K ¼ 521:3 K V 3 ¼ M3

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi γRT3 ¼ 0:72  1:4  287  521:3 ¼ 329:0 m=s

This value of V3 is the same as ðVsr  Vc Þ, where Vc ¼ 0 as required by the □ boundary condition due to reflection.

11.5 Isentropic One-Dimensional Waves of Finite Amplitude and Characteristics Small perturbation theory was applied to one-dimensional acoustic wave propagation in Chapter 9, where the perturbation is negligibly small so that the entire induced flow field is isentropic. In this section, waves causing finite, but weak, changes are of concern. Examples of finite wave include the moving shock wave discussed in the previous section. A rarefaction wave causing significant density decrease is also a finite wave. Due to the finite change in temperature, the acoustic speed changes and the wave propagation and interaction become nonlinear, unlike in Chapter 9, where a constant acoustic speed is used throughout the entire flow field and the phenomenon is linear. On the other hand, the complex wave interaction makes tracking local acoustic speed and particle velocity a complicated and tedious task; for example, if one follows the methods described in Sections 11.2 and 11.3, accounting for the relative

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11.5 Isentropic Waves of Finite Amplitude

289

speeds due to multiple passing of waves by using the stationary shock theory (i.e., moving waves with coordinate transformation) might be daunting. Here the finite changes are assumed to be still small so that the flow field is considered isentropic and the method of characteristics (MOC) can be extended for solving complex nonlinear propagation and interactions of waves (of the same of opposite families). The following describes the development of MOC for nonlinear wave propagation, providing an algorithmic approach to the problem. The extent to which the isentropic assumption is suitable will also be discussed. In Section 9.6, acoustic wave propagation equation and its general solution are, respectively, ∂2   a21 r2  ¼ 0 ∂t2

(9.49) (11.19)

 ¼ F ð x  a1 t Þ þ G ð x þ a1 t Þ

(9.50) (11.20)

where  is the perturbation velocity potential. The functions F ð x  a1 tÞ and Gð x þ a1 tÞ are constant for constant values of ð x  a1 tÞ and ð x þ a1 tÞ, respectively. Again, lines corresponding to ð x  a1 tÞ = constant and ð x þ a1 tÞ = constant are called characteristic lines or simply characteristics on the x-t plane described by dx=dt ¼ a1 . Equation (11.19) is derived using small-perturbation theory and is thus linear and, as a consequence, both F ð x  a1 tÞ and Gð x þ a1 tÞ are its solutions and so are their linear combinations shown by Eqn. (11.20) is an example. It is also noted from Eqns. (11.19) and (11.20) that due to the small amplitudes (of ρ0 , p0 , 0 T and v~0 ) and linearization, the acoustic wave possesses the following features: (1) it travels at a constant wave velocity a1 , (2) the wave shape is permanent, and (3) they are isentropic and linear waves. The continuity and momentum equations for finite waves can be derived from the more general form given in Chapter 8:
 ∂ρ
 ∂ρ ~ ¼ ~þ V ~ ⋅r ρ ¼ 0 þ r ⋅ ρV þ ρr ⋅ V ∂t ∂t

ρ

 ~
∂V ~ ⋅r V ~ ¼ rp þ ρV ∂t

(8.1d) (11.21)

(8.2e) (11.22)

These equations are nonlinear and finite waves are also nonlinear. The technique for solving these equations is the method of characteristics, as detailed in the following. For one-dimensional flow, Eqns. (11.21) and (11.22) are reduced, respectively, to ∂ρ ∂ρ ∂u þu þρ ¼0 ∂t ∂x ∂x

(11.23)

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290

11 Unsteady One-Dimensional Flows and Nonlinear Waves

∂u ∂u 1 ∂p þu þ ¼0 ∂t ∂x ρ ∂x

(11.24)

Assuming the medium ahead of the wave is quiescent, r ⋅ s ¼ 0 and dp ¼ a2 dρ. (Note that entropy changes due to the passing of finite waves.) A flow in a region with r ⋅ s ¼ 0 is homentropic. Then Eqn. (11.23) becomes

1 ∂p ∂p ∂u þu ¼0 (11.25) þρ a2 ∂t ∂x ∂x Multiplying Eqn. (11.25) by a and dividing by ρ and then by adding/subtracting the result to/from Eqn. (11.24) yields



∂u 1 ∂p ∂u 1 ∂p   þ ðu  aÞ ¼0 (11.26) ∂t ρa ∂t ∂x ρa ∂x The general solution to the nonlinear Eqn. (11.26) is difficult to obtain. However, for the observer “riding” on the particle, i.e., moving with ðu þ aÞ, dx ¼ ðu þ aÞdt or

dx ¼uþa dt

Substituting this expression into the exact differential of u leads to   ∂u ∂u ∂u ∂u ∂u ∂u du ≡ dt þ dx ¼ dt þ ðu þ aÞ dt ¼ þ ð u þ aÞ dt ∂t ∂x ∂t ∂x ∂t ∂x

(11.27)

(11.28)

Similarly, 

 ∂p ∂p þ ð u þ aÞ dp ¼ dt ∂t ∂x

(11.29)

By multiplying Eqn. (11.26) by dt, followed by invoking Eqns. (11.28) and (11.29), one finds du þ

dp ¼0 ρa

(11.30)

Similarly for the observer moving with ðu  aÞ, du 

dp ¼0 ρa

(11.31)

The reader is reminded of the resemblance that Eqns. (11.30) and (11.31) bear with the results of Problem 9.8 for the linear acoustic wave, where 0

u ¼

0

p ðthe  signs are for right- and left-running linear acoustic waves; ρ1 a1

respectivelyÞ

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11.5 Isentropic Waves of Finite Amplitude

291

t dx = u–a dt dp Compatibility : du – =0 ρa ) J– = u – γ2a –1 )

C– characteristic :

Particle path (pp):

dx =u dt

dx = u+a dt dp Compatibility : du + =0 ρa ) J+ = u + γ2a –1 )

C+ characteristic : t1

x1

x

Figure 11.4 Schematic showing the characteristic lines, with their slopes, compatibility conditions, and particle path affected by propagation of a finite-amplitude, nonlinear wave.

The above derivation reduces the governing partial differential equations, Eqn. (11.23) and (11.24) and their combined form, Eqn. (11.26), to ordinary differential equations, Eqns. (11.30) and (11.31). The requirement for such a reduction is that the rate of changes are observed by riding on particles moving with velocities, ðu  aÞ. The trajectory or these particles on an x-t plot is shown in Fig. 11.4. Equations (11.30) and (11.31) hold only along the following characteristic lines: Cþ characteristic line :

dx ¼uþa dt

C characteristic line :

dx ¼ua dt

Equations (11.30) and (11.31) are called the compatibility equations along the Cþ and C characteristic lines, respectively. It is noted that the characteristic lines shown in Fig. 11.4 are in general not straight lines, owing to the interaction between the finite-amplitude waves passing each other, resulting in changes in both u and a. In contrast, characteristics of linear acoustic waves described in Chapter 9 are straight lines with dx=dt ¼ a1 ¼ constant. Integrating Eqn. (11.30) along a Cþ characteristic yields ð dp ¼ u þ F ¼ const: ¼ J þ uþ (11.32) ρa The r ⋅ s ¼ 0 condition allows that a2 ¼ γRT ¼ γp=ρ and ρ ¼ γp=a2 . It can be seen that in Eqn. (11.32) F ¼ f ð p; ρÞ is itself a thermodynamic property. Following Eqn. (4.4),

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292

11 Unsteady One-Dimensional Flows and Nonlinear Waves

p ¼ c1 T γ=ðγ1Þ ¼ c2 a2γ=ðγ1Þ where c1 and c2 are constants. Therefore,

2γ dp ¼ c2 aðγþ1Þ=ðγ1Þ da γ1

(11.33)

(11.34)

and ρ ¼ c2 γa2=ðγ1Þ

(11.35)

Substituting Eqns. (11.34) and (11.35) into Eqn. (11.32) and integrating leads to uþ

2a ¼ const: ¼ J þ ðalong a Cþ characteristicÞ γ1

(11.36a)

u

2a ¼ const: ¼ J  ðalong a C characteristicÞ γ1

(11.36b)

Similarly

J þ and J  in Eqns. (11.36a) and (11.36b) are called the Riemann invariants for constant values of γ (γ is assumed to be constant during the integration process). Therefore, a characteristic carries on it information that relates not only u and a but also changes in them. It is left as an exercise to show that ð dp 2a ¼ (11.37) F≡ ρa γ  1 Therefore, along the a Cþ and a C characteristics (i.e., for an observer riding with the wave propagation speeds ðu þ aÞ and ðu  aÞ), respectively,   Dþ ∂ ∂ þ ð u þ aÞ ðu þ F Þ ≡ ðu þ F Þ ¼ 0 ðalong a Cþ characteristicÞ (11.38a) ∂t ∂x Dt   D ∂ ∂ þ ðu  aÞ ðu þ F Þ ≡ ðu  F Þ ¼ 0 ðalong a C characteristicÞ (11.38b) ∂t ∂x Dt where D =Dt represent the rate of change to the observer riding on the wave, or traveling with a velocity of ðu þ aÞ and ðu  aÞ, respectively. Such rates of change are analogous to the material or substantial derivatives in general fluid flow, i.e., D=Dt ≡ ½∂=∂t þ uð∂=∂xÞ. Equations (11.38a) and (11.38b) indicate that ðu  F Þ remains unchanged on each of C characteristics, respectively. In the limit of waves with negligible amplitude (u ≪ a), i.e., linear acoustic waves, D ∂ ∂ ¼ a ∂x Dt ∂t

(11.39)

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11.6 Types of Waves and Interactions

293

with general solutions having the functional forms of F ð x  a1 tÞ or Gð x þ a1 tÞ, as Eqn. (9.50) has demonstrated. Nonlinear (finite-amplitude) waves thus carry information at a velocity of ðu  aÞ, as opposed to the acoustic velocity a of linear waves. It is now possible to solve Eqns. (11.35) and (11.36) for u and a: a¼

γ1 þ ðJ  J  Þ 4

1 u ¼ ðJ þ þ J  Þ 2

(11.40)

(11.41)

Thus at any given point on the x-t plane, u and a can be solved in Eqns. (11.40) and (11.41) given known values of J þ and J  . The solution is schematically shown in Fig. 11.4, where the solution leading to the particle path is also shown by the curve with dx=dt ¼ u. The particle velocity u is due to the induced mass motion, similar to Vb in Sections 11.2 and 11.3.

11.6 Types of Waves and Interactions Several observations of the current unsteady nonlinear wave propagation can be made. First, the calculation procedure is similar to that using the method of characteristics for two-dimensional isentropic (and irrotational) flows discussed in Chapter 10. While the Riemann invariants in Chapter 10 are ð þ δÞ and ð  δÞ, they are now J þ and J  . The data curve or the initial value line (as in Chapter 10) will necessarily provide the initial data for calculations further along the characteristics and into the t-x space, as demonstrated later by Example 11.6. Secondly, the simple (wave) region exists where there is no interaction between waves, and a simple wave is one that propagates into a region of uniform flow (or no flow at all). In the simple wave region the flow is determined by either J þ or J  , depending on the direction of propagation. As a consequence, both u and a are constant along a given characteristic and the characteristic is a straight line. An example of a simple wave region is given in Fig. 11.5a, which shows disturbances generated at some point in a uniform flow, with a speed of uo , in a long tube. To the laboratory observer the disturbances are carried by the wave traveling at a speed of ðu þ aÞ with the leading wave front at a speed of ðuo þ ao Þ. Note that the acoustic speed may vary with time (e.g., a sinusoidal disturbance creates alternatively compression and expansion waves causing an increase or decrease in the acoustic speed). resulting in different slopes for the Cþ characteristics on the t-x plane, shown in Fig. 11.5b. Along each of the Cþ characteristics (propagating in the +x direction, or right-running) is a respective constant J þ ¼ u þ 2a=ðγ  1Þ ¼ constant, and the constant varies from one to the other characteristic. On the other hand, all the C characteristics (propagating in the –x direction, or left-running) have the same constant for J  (¼ uo  2ao =ðγ  1Þ ¼ u  2a=ðγ  1Þ) as they originate from the uniform flow region where both u and a are constants. An infinite number of C

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294

11 Unsteady One-Dimensional Flows and Nonlinear Waves (a) Traveling waves Source of

Uniform flow region u0

* disturbance

p0, ρ0, T0, a0

x (b)

t

C+ characteristics (J+ = different constant along different characteristics) Uniform region: u0 = constant a0 = constant C– characteristics (J– = u0 −

za0 = constant) γ –1 x

Figure 11.5 (a) Generation and propagation of a simple wave; (b) characteristics on the t-x plane with their Riemann invariants.

characteristics can be drawn on the t-x plane. Thus one can conclude that J  ¼ uo  2ao =ðγ  1Þ ¼ constant everywhere and that the flow is determined by J þ ). Third, in a non-simple (wave) region, wave interaction (passing each other) takes place. An example is shown in Fig. 11.6a, where a diaphragm divides a highand low-pressure zones in a tube with both ends closed, and the compression and expansion waves are generated after the rupture of the diaphragm (Fig. 11.6b), Sometime later the reflected waves from the end walls (Fig. 11.6c) pass through each other (Fig. 11.6d). Figure 11.6d shows the non-simple wave region, where the right- and left-running waves (corresponding to the reflected expansion and compression waves, respectively) intersect. At the point of intersection, the flow is determined by both J þ or J  . One can further see the complexity of interaction by extending the event of Fig. 11.6d to a later moment when the left-running compression wave is reflected from the left wall to become right-running and meet with the left-running expansion wave reflected from the right end wall (Fig. 11.6e). By this time, the compression wave has been weakened while the expansion wave also has a diminished rarefying capability, resulting from their encounter (Fig. 11.6d) with the other. For a more quantitative description of wave interactions, consider the example shown in Fig. 11.7a. In Fig. 11.7, compression and expansion waves interact due to the action of piston 1 continuously accelerating to the right to reach a constant speedup , generating a series of compression waves that travels to the right and piston 2 also continuously accelerating to the right to reach a constant speed up =2, generating a series a series of expansion waves that travels to the left. For illustration, assume that no reflected wave from either piston surface has yet occurred and that the two pistons reach their respective steady-state speed at the same time. Consider three characteristics generated by each piston action at three different instants: at

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11.6 Types of Waves and Interactions

295

Diaphragm/Initial contact surface (a) p1

(b)

Expansion wave

p2 < p1

Contact surface

Shock wave

p2

p1

Reflected expansion wave

Reflected shock wave Contact surface

(c)

Shock-expansion interaction

Contact surface

(d)

(e)

Figure 11.6 (a) A diaphragm separating high- and low-pressure gases in a tube; (b) after the rupture of the diaphragm, a normal shock wave forms and propagates into the low-pressure gas while an expansion forms and propagates into the high-pressure gas, with the contact surface being carried along by the shock wave (or being repulsed away from the expansion wave); (c) reflected normal shock and expansion waves; (d) interacting shock and expansion waves, forming non-simple wave region; (e) impending second interaction of shock and expansion waves.

the beginning, in the middle, and at the end of acceleration. The three right-running characteristics generated by piston 1 emanate from points, A, B, and C on the t-x plane, and the three left-running characteristics by piston from point D, E, and F. Several qualitative features of Fig. 11.7b are summarized in the following, which also help to point out the need of repeatedly using the method of characteristics for quantitative analysis of the flow field and wave propagation. (1) Because of the compression action of piston 1, the slopes of the three characteristics prior to interaction are such that ðdx=dtÞC3 > ðdx=dtÞB2 > ðdx=dtÞA1 . The slopes of these right-running characteristics are ðdx=dtÞ ¼ ðu þ aÞ. Equation (11.33), p ¼ c1 T γ=ðγ1Þ ¼ c2 a2γ=ðγ1Þ , indicates increases in T and thus in a on the characteristic generated following characteristic A-1. The induced

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296

11 Unsteady One-Dimensional Flows and Nonlinear Waves Piston 1

Piston 2

Accelerating to up, generating right-running compression waves

Accelerating to up/2 generating left-running expansion waves

t

dx

=

up

9

8

dt

3

6

7 5

2

F (up /2)

4

B

(u

p /2

)

C (up )

=

u p1

Simple wave region 2a0 J+ = u0 + γ−1 up 2a4 = + 4 γ−1 = ... = Constant dx = up2 = up/2 dt

1

E (up /4)

D

A

Particle path x

Simple wave region 2a0 J− = u0 − γ−1 up 2a2 = − 2 γ−1 = ... = Constant

Figure 11.7 (a) Schematic of the relative motion of two pistons in a long tube; (b) characteristic waves formed due to piston 1 moving with a constant speed up and piston 2 with up =2, both to the right; also included is an example of particle path. The grid numbering is used in Example 11.5.

mass motion also increases from A-1 to C-3 due to the increased piston speed. Therefore, the slope increases from A-1 to C-3 due to (1) an increase in u by the increased piston speed and (2) an increase in a resulting from compression. (2) Similar reasoning can be made for the left-running characteristics. These characteristics have slopes ðdx=dtÞ ¼ ðu  aÞ with magnitudes in the order of jdx=dtjD1 > jdx=dtjE4 > jdx=dtjF7 . Such an order arises due to two factors: (1) an increase in uin the positive x direction by the piston motion and (2) decreases in T and thus in a on characteristics from D-1 to F-7 caused by the passing expansion/rarefaction waves. (3) It is noted that the regions A-1–2-3-C-B-A and D-E-F-7-4-1-D on the t-x plane are simple wave regions, where the qualitative trend of the slope of a characteristic can be easily explained as the slope changes due to either compression or expansion/rarefaction. The region A-D-1-A is the undisturbed region. The qualitative trend of the slopes of the rest of the characteristic line

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11.6 Types of Waves and Interactions

297

segments (1–4, 1–2, 2–3, 3–6, 4–5, . . ., etc.) cannot be easily determined as they are now in the non-simple wave region, where Cþ and C characteristics intersect. They will have to be determined by repeatedly using the method of characteristics. In the case depicted by Fig. 11.7, the relative magnitude of their slopes depend on the relative piston speeds of ðdx=dtÞp1 and ðdx=dtÞp2 . (4) Whether in simple or non-simple wave regions, the local magnitudes of slopes, ðu þ aÞ and ðu  aÞ, (in other words, u and a) are needed to determine the location of wave intersection and to construct the network of characteristics. The following example illustrates the successive use of the method of characteristics for quantitative results of the flow field and wave propagation. Find the slopes of line segments A-1, B-2, D-1, E-4, 1–2, and 1–4 in Fig. 11.7b (i.e., find the gas and acoustic velocities at each of the intersection locations of characteristics). Assume that both pistons are continuously moved to reach their respective steady-state speed over the same period of time that is sufficiently short so that no reflected waves from each piston have occurred.

EXAMPLE 11.5

Solutions – The gas between the two piston is initially at rest; therefore uo ¼ 0 and a ¼ ao . Lines A-1

Riemann Invariants þ

J ¼ uo þ 

2ao γ1 2ao γ1

¼ u1 þ

D-1

J ¼ uo 

A-B

2aA 2aB J  ¼ uA  γ1 ¼ uB  γ1

B-2

2aB 2a2 J þ ¼ uB þ γ1 ¼ u2 þ γ1

1–2

2a1 2a2 J  ¼ u1  γ1 ¼ u2  γ1

B-C

2aC 2aB J  ¼ uB  γ1 ¼ uC  γ1

C-3

2aC 2a3 J þ ¼ uC þ γ1 ¼ u3 þ γ1

2–3

2a3 2a2 J  ¼ u2  γ1 ¼ u3  γ1

D-E

2aE D J þ ¼ uD þ 2a γ1 ¼ uE þ γ1 þ

¼ u1 

2a1 γ1 2a1 γ1

1–4

J ¼ u1 þ

E-4

2a4 2aE J  ¼ uE  γ1 ¼ u4  γ1

2–5

2a5 2a2 J þ ¼ u2 þ γ1 ¼ u5 þ γ1

4–5

2a5 2a4 J  ¼ u4  γ1 ¼ u5  γ1

3–6

2a6 2a3 J þ ¼ u3 þ γ1 ¼ u6 þ γ1

2a1 γ1

¼ u4 þ

2a4 γ1

Known

Solutions

uo ¼ 0 ao

u1 ¼ uo ¼ 0 a1 ¼ ao

uA ¼ 0 u uB ¼ 2p aA ¼ ao

aB ¼ ao þ γ1 4 up u2 ¼ 12 up a2 ¼ ao þ γ1 4 up

Slope dx dt

¼ u1 þ a1 ¼ ao

dt D1

¼ u1  a1 ¼ ao

dxA1

dx dt B2

dx dt 12

uC ¼ up

aC ¼ ao þ γ1 2 up u3 ¼ up a3 ¼ ao þ γ1 2 up

dx dt C3

dx dt 23

uD ¼ 0 u uE ¼ 4p aD ¼ ao

aE ¼ ao  γ1 8 up u4 ¼ 14 up a4 ¼ ao  γ1 8 up

dx dt 14

dx dt E4

u5 ¼ 34 up a5 ¼ ao þ γ1 8 up u6 ¼ 54 up Þ a6 ¼ ao þ 3ðγ1 8 up

dx

¼ u2 þ a2 ¼ ao þ γþ1 4 up ¼ u2  a2 ¼ ao þ 3γ 4 up

¼ u3 þ a3 ¼ ao þ 1þγ 2 up ¼ u2  a3 ¼ ao þ 3γ 2 up

¼ u4 þ a4 ¼ ao þ 3γ 8 up ¼ u4  a4 ¼ ao þ γþ1 8 up

dt

¼ u5 þ a5 ¼ ao þ γþ5 8 up

dt 45

¼ u5  a5 ¼ ao þ 7γ 8 up

dx25 dx dt 36

¼ ao þ

h¼ u6 iþ a6 3γþ7 8 up (continued)

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298

11 Unsteady One-Dimensional Flows and Nonlinear Waves

(continued) Lines

Riemann Invariants

5–6

2a5 2a6 J  ¼ u5  γ1 ¼ u6  γ1

F-F

2aE 2aF J þ ¼ uE þ γ1 ¼ uF þ γ1

4–7 F-7

þ

J ¼ u4 þ

2a4 γ1

¼ u7 þ

Known

2a7 γ1

2a5 2a8 J þ ¼ u5 þ γ1 ¼ u8 þ γ1

7–8

2a8 2a7 J  ¼ u7  γ1 ¼ u8  γ1

6–9

2a6 2a9 J þ ¼ u6 þ γ1 ¼ u9 þ γ1 2a8 2a9 J  ¼ u8  γ1 ¼ u9  γ1

Slope dx

dt 56 ¼hu6  ia6 up ¼ ao þ 133γ 8

uF ¼

up 2

aF ¼ ao  γ1 4 up u7 ¼ 12 up a7 ¼ ao  γ1 4 up

2a7 2aF J  ¼ uF  γ1 ¼ u7  γ1

5–8

8–9

Solutions

dx dt 47

¼ h u7i þ a7 3γ 4 up

dt 47

¼ h u7i þ a7 γþ1 4 up

¼ ao þ dx ¼ ao þ

u8 ¼ up a8 ¼ ao

u9 ¼ 32 up Þ a9 ¼ ao þ ðγ1 4 up

dx

dt 58 ¼ u8 þ a8 ¼ ao þ up dx dt 78 ¼ u8  a8 ¼ ao þ up

dx dt 69

¼ ao þ dx

¼ h u9i þ a9 7γ 4 up

i a9 dt 69 ¼hu9  ¼ ao þ γþ5 4 up

Comments – It is interesting to note that u8 ¼ up and a8 ¼ ao . At point 8, the intersecting J þ and J  characteristics originate from points B and F, respectively, where the amount of piston advancing and withdrawing are equal in magnitude (¼ up =2) and cancel out each other. Because the difference in the piston advancing speed for the points of interest has the unit of up =8, one can see that variations in the acoustic and gas velocity are multiples of ðγ  1Þup =8 and up =8, respectively. The increment in the acoustic velocity is smaller than that in the gas velocity by □ a factor of (γ  1). It is useful to revisit the qualitative trend regarding the slope of characteristics, especially those in the non-simple wave region, based on the quantitative results of Example 11.5. The following observation can be made. (1) A left-running characteristic (i.e., of the C family) passing through successive Cþ characteristics (here the compression waves) experiences increases in both u and a. Consider for example the characteristic D-1-2-3 that experiences three compression waves emanating from Points A, B, and C. As shown in Example 11.5, u1 < u2 < u3 and a1 < a2 < a3 . The increase in u from point 1 to point 3 is due to the action of piston 1. As comments of Example 11.5 indicate, the increase in a from point 1 to point 3 is smaller than that in gas velocity u. Consequently, the absolute values of slopes of these characteristic line segment are such that jdx=dtjD1 < jdx=dtj12 < jdx=dtj23 . Similar qualitative observations can be made for the other two characteristics of the C family: E-4-5-6 and F-7-8-9. Therefore, the left-running (C ) characteristics accelerate as they propagate into the gas being compressed by the motion of piston 1. Downloaded from https:/www.cambridge.org/core. Columbia University Libraries, on 24 Jun 2017 at 04:20:25, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/9781316014288.012

11.6 Types of Waves and Interactions

299

(2) A right-running characteristic (i.e., of the Cþ family) passing through successive C characteristics experiences increases in u and decreases in a. Consider for example the characteristic A-1-4-7 that experiences three rarefaction waves emanating from Points D, E, and F. As shown in Example 11.5, u1 < u4 < u7 and a1 > a4 > a7 . Therefore, jdx=dtjA1 < jdx=dtj14 < jdx=dtj47 . Similar qualitative observations can be made for the other two characteristics of the Cþ family: B-2-5-8 and C-3-6-9. One can conclude that the right-running (Cþ ) characteristics decelerate as they propagate into the gas being rarefied by the motion of piston 2. (3) To improve the accuracy and resolution of the flow region, more characteristics can be added. The number of both Cþ and C characteristics can be increased to a number n by, for example, choosing characteristics with velocity increments of up =n and up =2n, respectively, on the advancing piston 1 and withdrawing piston 2 (so that there are ðn þ 1Þ characteristics for each of the Cþ and C families). (4) The reader is encouraged to make observations for the scenario where the advancing motion is twice as weak as the withdrawing motion (up =2 vs. up ), which is left as an exercise (Problem 11.5). It is again noted that for the simple wave region A-1-2-3-C-B-A shown in Fig. 11.7b, the flow is determined by the Riemann invariant J þ , because J  is uniformly constant in the region it originates from the uniform region A-D-1-A where both the flow and acoustic velocity are constant (uo and ao , respectively). The following example demonstrates how the flow can be determined in the simple wave regions. EXAMPLE 11.6

Determine the acoustic velocities at points B, C, E, and F in

Fig. 11.7b. Solutions – Consider a C characteristic that originates from the uniform region and intersects the surface of piston 1 at point B and C. Then J  ¼ uo 

2ao 2aB ¼ const: ¼ uB  γ1 γ1

u

Because uo ¼ 0 and uB ¼ 2p , aB ¼ ao þ

γ1 up 4

Similarly, for point C J  ¼ uo 

2ao 2aC ¼ const: ¼ uC  γ1 γ1

The fact that uC ¼ up leads to aC ¼ ao þ

γ1 up 2

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300

11 Unsteady One-Dimensional Flows and Nonlinear Waves

Now consider a Cþ characteristic that originates from the uniform (undisturbed) region and intersects the surface of piston 2 at point E and F. Then J þ ¼ uo þ Because uo ¼ 0 and uE ¼

2ao 2aE ¼ const: ¼ uE þ γ1 γ1

up 4

aE ¼ ao 

γ1 up 8

For point F, J þ ¼ uo þ With uo ¼ 0 and uF ¼

2ao 2aF ¼ const: ¼ uF þ γ1 γ1

up 2

aF ¼ ao 

γ1 up 4

Comments – These acoustic velocities determined in Example 11.6 using the simple wave theory are identical with those obtained in Example 11.5 assuming that the piston path line on the t-x plane is a characteristic. For each of the simple wave regions, A-1-2-3-C-B-A and D-E-F-7-4-1-D, one can draw multiple characteristics originating from the uniform flow region. These characteristics reach the surface of the two pistons. Each of the Riemann invariants is therefore constant, J  and J þ , respectively, for A-1-2-3-C-B-A and D-E-F-7-4-1-D. Thus, the piston path line is □ a characteristic, when it is still in the simple wave region.

11.7 Centered Expansion In Fig. 11.8a, the expansion/rarefaction is caused by a continuously withdrawn piston in a long tube filled with initially quiescent gas (similar to Piston 2 described in Fig. 11.7), where the Cþ characteristics emanate from the curved t-x path line of the piston. Imagine that the piston motion is impulsively achieved, the curvature of the piston path line is zero and all the characteristics emanate (or rather for this case, radiate) from a single point, as shown in Fig. 11.8b. This idealized wave system is used to illustrate the variation of the flow and gas properties throughout the field. For the impulsively withdrawn piston motion described in Fig. 11.8b, let its location be Xp ðtÞ such that dXp ¼ up ð< 0 in Fig: 11:8Þ dt The piston motion depicted in Fig. 11.8 generates right-running, Cþ -family characteristics that form a centered expansion fan. By setting the velocity increment

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11.7 Centered Expansion

301

(a)

up

u0 = 0 a0

(n+1) characteristics in simple-wave region

(b) . Xp(t) = up

Cn+ t

C+n–1 C+n–2

Pa pat r ticle h

Uniform region

J− = Constant 2a =u+ γ−1 2a0 C+2 = γ−1 C+1 Throughout all the C+0 flow regions

Undisturbed region u=0 a = a0

0 xn x0

x

Figure 11.8 (a) Schematic of piston motion with a constant speed away from the initially quiescent gas in a long tube, generating centered expansion/rarefaction waves into the gas; (b) characteristic waves formed due to the piston moving with a constant speed up ; also included are two examples of particle path.

between any two adjacent characteristics to be up =n, there are ðn þ 1Þ characteristics with the leading and trailing ones denoted by C0þ and Cnþ , respectively. Ahead of the C0þ characteristic (i.e., x=ao t ≥ 1) is the undisturbed region where u ¼ 0 and a ¼ ao . Downstream of the Cnþ characteristic (i.e., x=ao t ≥ up =ao ) is the uniform region where the gas has attained the velocity of the piston, that is, on and downstream of the Cnþ characteristic u ¼ up . Between these two regions lies the expansion fan comprised of simple waves, whose fronts are represented by C0þ , C1þ , . . . Cnþ in Fig. 11.8b. Recall that on each characteristic, the local gas velocity u ¼ constant and the local acoustic speed a ¼ constant. Therefore, each of these Cþ characteristics is a straight line, as dx=dt ¼ ðu þ aÞ ¼ constant. Because of the simple-wave nature, J ¼ u 

2a 2ao ¼ constant ¼ uo  ðthroughout the entire flow regionÞ γ1 γ1

With this constant value of J  , the acoustic speed on each characteristic can be calculated. For example on the Cnþ characteristic, un ¼ up leads to an ¼ ½ao þ ðγ  1Þup =2. In general, on the ith characteristic

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11 Unsteady One-Dimensional Flows and Nonlinear Waves

ui ¼

iup n

(11.42a)

Upon substituting Eqn. (11.42a) into the above equation for J  , one obtains ai ¼ ao þ

dx dt

iðγ  1Þ up ði ¼ 0; 1; 2; …; nÞ 2n

¼ ui þ ai ¼ ao þ i

(11.42b)

i ð γ þ 1Þ up 2n

Eqn. (11.42) provides flow condition along characteristics. For a given n the flow field is determined with the associated resolution and accuracy. One the other hand, it is desirable to know the flow and acoustic velocities at an arbitrary point (x, t) on the t-x plane. The procedure briefly discussed so far in this section leads to: u ¼0 ao a x ≥ 1 ði:e:; the undisturbed regionÞ ¼ 1 for ao ao t

(11.43)

The Cnþ characteristic provides the boundary for the uniform region, so that

  dx x ð γ  1Þ γþ1 dx ¼ ¼ up þ a ¼ up þ ao þ up ¼ a o þ up ¼ dt t 2 2 dt n The simple wave (i.e., the fan) region is therefore defined by the following expression: 1≥

up γ þ 1 up x iðγ  1Þ up ≥ þ and ai ¼ ao þ ao t ao 2 ao 2n

In the uniform region, Eqn. (11.42) yields up u ¼ a o ao up a γ  1 up γ þ 1 up x ≥ ði:e:; the uniform regionÞ ¼1þ for 1 þ ≥ ao 2 ao 2 ao ao t a o

(11.43)

In the fan (i.e., simple wave) region,   dx x ð γ  1Þ ð γ þ 1Þ ¼ ¼ up þ a ¼ up þ ao þ u p ¼ ao þ up dt t 2 2 Solving for u in terms of x and t and then substituting the result in solution for a yields

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11.7 Centered Expansion



u 2 x 1 ¼ ao γþ1 ao t

303

(11.44a)



a γ1 x x γ þ 1 up 1 ≥1 þ ¼1 ði:e:; the fan regionÞ (11.44b) for 1 ≥ ao γþ1 ao t ao t 2 ao The derivation of these two expressions is left as an exercise (Problem 2.27). The velocity thus obtained is now used to determine the particle trajectory (PT), XPT , as a function of time. Attention is paid to the fan region, as the particle trajectory in the other two regions is readily known due to the uniform and undisturbed conditions there. Because the characteristic is a straight line, Eqn. (11.44a) can be written as

x dx 2ao 2 x or (11.45) þ ¼ u¼ t dt γþ1 γþ1t which is a linear differential solution. One may choose to find a particular solution for dx x 2ao 2 x ¼ ¼ þ dt t γþ1 γþ1t

(*)

and a general solution for dx 2 x ¼ dt γ þ 1 t By solving the above expression (*) for x=t, the particular solution in the fan region is x 2ao x γ þ 1 up ¼ for 1 ≥ ≥1 þ ði:e:; the fan regionÞ t ao t 2 ao γ1 while the general solution is x ¼ Kt2=ðγþ1Þ with K being a constant to be determined. Combining the particular and general solutions yields x¼

2 ao t þ Kt2=ðγþ1Þ γ1

(11.46)

For the initial condition, assume that the expansion wave passes a particle located at xo (=ao to ) with u ¼ 0 at t ¼ to for t ≤ to . Differentiating both sides of Eqn. (11.46) and setting the result equal to zero gives

dx 2 2 ð1γÞ=ðγþ1Þ ao þ K t ¼0¼ uðx0 ; t0 Þ ¼ dt x0 ; t0 γ1 γþ1 o

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11 Unsteady One-Dimensional Flows and Nonlinear Waves

Therefore, K¼

γ þ 1 ðγþ1Þ=ð1γÞ ao t γ1 o

Substituting this expression into Eqn. (11.46) and rearranging the result yields 2=ðγþ1Þ 2 γþ1 t ao t þ xo x¼ γ1 γ1 to

(11.47)

ð1γÞ=ðγþ1Þ x 2 γ þ 1 xo t þ ¼ ao t γ  1 γ  1 ao t o t o

(11.48)

Define non-dimensional parameters e x and et as e x≡

x t ; et ≡ ao t to

(11.49)

Then Eqn. (11.47) becomes e x¼

2 γþ1 2 γ þ 1 ð1γÞ=ðγþ1Þ ð1γÞ=ðγþ1Þ et e þ þ ¼ x oet γ1 γ1 γ1 γ1

(11.50)

where e x o ¼ xo =ao to ¼ 1: Eqn. (11.44) can be rewritten as e u¼

e a ¼1

2 ð1  e xÞ γþ1

γ1 γþ1 e ð1  e x Þ for 1 ≤ e x≤1 þ u p ði:e:; the fan regionÞ γþ1 2

(11.51)

where e u≡

up u a ; e up ≡ ; e a≡ ao ao ao

(11.52)

Examining Eqns. (11.50) and (11.51) suggests that e x ¼ x=ao t and et ¼ t=to are the similarity variables for the one-dimensional unsteady flow induced by the centered expansion wave. The self-similar nature of the gas motion is clearly seen in the wave diagram of Fig. 11.8. The relevant time scale is to and the lack of a geometric dimension in the direction of motion necessitates the use of ao t as the unique and relevant length scale. Such a lack is also seen in that as t increases, the fan occupies an increasingly large extent while keeping a fixed fan angle. The spread of the fan arises from the decrease in acoustic speed of successive waves as time passes. It is of interest to check the validity of Eqn. (11.50). Consider the beginning of x¼e x o ¼ 1 for et ¼ 1 the expansion at t ¼ to , when et ¼ 1 and x ¼ xo ¼ ao to . That is, e and such a condition is indeed satisfied by Eqn. (11.50).

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11.7 Centered Expansion

305

Referring to Fig. 11.8b, several results of the gas particle motion within the expansion fan can be observed and summarized in the following. These results are left as exercise problems: (1) As successive expansion wave fronts pass the particle, the location changes from e e x ¼e x o ¼ 1 upon encountering C0þ characteristic to e x¼e x n ¼ 1 þ γþ1 2 u p at the þ passing of the Cn characteristic. Therefore, the total displacement of the particle is Δe x ¼e xn  e xo ¼

γþ1 e up 2

(11.53)

That is, the faster the piston motion, the larger the particle displacement. Because e u p < 0, the particle moves in the –x direction, which is consistent with the previous finding that an expansion wave (right-running in the +x direction, as in Fig. 11.8b) displaces particles in the direction opposite to its propagation. (2) There exists a piston speed that is equal to the local sonic speed, for which Eqn. (11.42b) yields

dx γþ1 up ¼ 0 ¼ u þ a ¼ ao þ dt Cþ 2 That is, the Cnþ characteristic becomes vertical in the t-x plane when jup jM¼1 ¼

2 ao γþ1

(11.54)

(3) To be carried out as an end-of-chapter exercise (Problem 11.6), the particle at xo spends a duration of  ð1þγÞ=ð1γÞ γ1 e e e e Δt ¼ t n  t o ¼ 1 þ 1 up 2

(11.55)

e p < 0 and ð1 þ γÞ=ð1  γÞ < 0, as the piston in the expansion fan. Because u speed increases, the time the particle spends inside the expansion fan increases. Such an increase is accompanied by the fact that the fan angle increases as the slope of the Cnþ characteristic decreases (dx=dt approaching zero in Fig. 11.8b). 2 ao ) the slope of the In fact, when the piston becomes supersonic (jup j > γþ1 þ Cn characteristic becomes negative on the t-x plane, which is not possible as discussed below in (4). (4) Because ð1 þ γÞ=ð1  γÞ < 0 and the value in the bracket of Eqn. (11.55) must be greater than zero for Δet > 0, therefore, the following condition must be satisfied: je upj ≤

2 2ao or jup j ≤ γ1 γ1

(11.56)

The less-than-or-equal sign in Eqn. (11.56) holds for the escape velocity defined as jup jescape ¼

2ao γ1

(11.57)

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306

11 Unsteady One-Dimensional Flows and Nonlinear Waves

with which the piston leaves a vacuum where no sound propagation is possible (and the temperature is absolute zero) and the gas can no longer sense the piston motion, and it separates or escapes from the piston surface. Therefore, it is not possible for the piston to be withdrawn at a supersonic speed. (5) Along the Cþ family of characteristics, which form the centered expansion fan in Fig. 11.8, Eqn. (11.30) indicates that du þ dp=ρa ¼ 0 on each characteristic. Because on a given characteristic, u ¼ constant p is also constant. The changes in u and p across the fan region are given by the changes on a C characteristic originating in the undisturbed region. For example, the change in u from the þ to Ciþ characteristic is equal to Δui : Ci1 Δui ≡ ui  ui1

(11.58)

Similarly, because along a C characteristic du ¼ dp=ρa, ði Δpi ≡ pi  pi1 ¼

ρadu

(11.59)

i1

The pressure change due to the fan is thus ðn Xn Xn ð i Δp ¼ ρadu ¼ Δp ¼ i i¼1 i¼1 0

ρadu

(11.60)

i1

Because both ρ and avary through the fan, piecewise integration is needed, for infinitesimal changes from the leading characteristic (C0þ ) to the trailing char0 acteristic (Cnþ ), Δp ¼ ρao u .

11.8 Approximation for Small-Amplitude Waves For the center expansion wave shown in Fig. 11.8, the limiting case where je u p j ! 0 or jup j ≪ ao constitutes the acoustic wave. The slope of the wave is

  dx γþ1 up ≈ a o ¼ ao þ (11.61) dt Cþ 2 for all characteristics (as for acoustic waves) and the expansion fan angle is negligible. (In Chapter 9, the acoustic speed in the undisturbed gas medium is denoted by a1 .) The duration for a particle under the influence of the expansion wave (or the time it takes for the particle to attain the piston speed after to ) is 

γ1 e Δet ¼ 1 þ up 2

ð1þγÞ=ð1γÞ

 1≈ 

γþ1 γ þ 1 xo up e u p or Δt ¼  2 2 a2o

(11.62)

while the last wave, the Cnþ characteristic, passes at

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11.9 Compression Wave and Shock Formation

t ¼ tn ¼ to þ Δt ¼



γ þ 1 up xo xo 1 ≈ ¼ to 2 ao ao ao

307

(11.63)

That is, the particle spends no (or negligible) time in the expansion fan (i.e., very thin wave). Equation (11.53) is used to obtain the displacement of a particle, which is

γ þ 1 up 2 xo Δx ≈ ðΔe x Þao to ¼  2 ao

(11.64)

 2 Equation (11.64) suggests that Δx due to the acoustic wave is of the order of up =ao and is thus negligible, unless for very large xo . Due to the negligible expansion fan angle, the particle must be very far away from the piston to experience an appreciable displacement.

11.9 Compression Wave and Shock Formation An advancing piston will lead to the collapse of compression waves, eventually forming a shock wave in a sufficiently long cylinder. Two types of piston motion are considered: the impulsively advanced and the gradually advanced, with the former being the limiting case of the latter. An impulsively advanced piston with dXp =dt ¼ up > 0, shown in Fig. 11.9, will necessarily cause a shock wave at the origin of motion. In Fig. 11.9, the region ahead of the compression wave is designated as region I, while that downstream of the wave is region II. Assuming that the gas in region I is initially quiescent, the following can be stated:

Along the CIþ characteristic :



dx 2aI ¼ aI and JIþ ¼ dt Cþ γ1

(11.65)

I

þ Along the CII characteristic :



dx dt

þ CII

¼ up þ aII and JIIþ ¼ up þ

As a result of compression up > 0 and aII > aI . Thus,



dx dx > dt Cþ dt Cþ II

2aII γ1

(11.66)

(11.67)

I

þ characteristics are steeper on the t-x plane and will intersect (i.e., Therefore the CII catch up) with the CIþ characteristics. Thompson (1972) describes details that lead to the following relationship

up 2 1 MIn  ¼ (11.68) MIn aI γ  1

where MIn is the Mach number of the shock propagating into region I. With MIn thus determined, normal shock relationships can be used to determine ratios of

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11 Unsteady One-Dimensional Flows and Nonlinear Waves

u=o a = ao p = po

. Xp = up

t Particle path

Figure 11.9 (a) Schematic of piston moving into the initially quiescent gas in a long tube, generating compression waves into the gas; (b) characteristic waves formed due to the piston moving with a constant speed up ; also included is an example of particle path.

II

J+

C+

=u p + 2a cha γ− II rac 1 ter isti cs

Piston path

(II) uniform region (I) undisturbed region

C+I characteristics

x

( dxdt (C = a = a = constant I

+ I

J+ = uI +

o

2aI 2ao = γ−1 γ−1

properties across the shock: pII =pI , TII =TI , and so on. It is noted that due to the nonisentropic and non-homentropic nature of the piston-induced shock flow field, the concept of Riemanian invariants is no longer valid. For example, there is no similar result for Δp as 24)shown by Eqn. (11.60) for expansion waves. Consider the general continuous piston motion prescribed by Xp ¼ Xp ðtÞ, depicted in Fig. 11.10 (adopted from Thompson, 1972) With acceleration of the piston motion, the later compression waves catch up with the earlier ones to cause a finite pressure increase across a collapsed, thin wave, thus forming a shock wave (this has been described in Section 4.3). The location of shock formation is designated as point a in Fig. 11.10., where the first two compression waves collapse. As the acceleration is increased, point a moves closer to the origin of motion, approaching the limiting case depicted in Fig. 11.9. Once the shock forms, the condition of homentropy ceases to be valid and the method of characteristics is no longer applicable. The Riemann invariant, J  , is only uniform on the t-x plane in the region bounded by the limiting C characteristic and the piston path line schematically shown in Fig. 11.10. The limiting C characteristic is the one in Fig. 11.10 that passes point a and still possesses the equality, JI ¼ JII . The C characteristic passing point b in Fig. 11.10 illustrates that downstream of the limiting C characteristic, the Riemann invariant J  assumes different values as it crosses from region (I) into

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11.9 Compression Wave and Shock Formation

309

. Xp = (t) = up (t)

t

Xp (t), piston path line C−II (J−II ≠ J−I ) C−

a

t = ta– t0 *

c

C+I

right-traveling or C+ shock

C+I Undisturbed region (I); u = 0, a = a0

e av tw

t0

C+II Typical right-running characteristics in region (II)

Uniform region (II) u≠0

b

ta

Particle path

C−I (JI− = −2a0/(r−11))

firs

xs

x Limiting C− characteristic (i.e., JI− = JII−)

Typical right-running characteristics in region (I)

Figure 11.10 (a) (a) Schematic of piston moving with varying speeds into the initially quiescent gas in a long tube, generating compression waves into the gas; (b) characteristic waves formed due to the piston moving with a speed up ðtÞ that increases with time. Note that CI is not possible downstream of x (corresponding to a time t ¼ ta  to ) as characteristics do not exist if rs ≠ 0.

region (II), with point b separating the two segments of the two characteristics CI  and CII . In region (I), JI ¼ 2ao =ðγ  1Þ while in region (II), JII ¼ uII  2aII =ðγ  1Þ ¼ up  2aII =ðγ  1Þ. The location of shock formation (point a in Fig. 11.10, with location denoted by xs ) is of interest, and so is the associated time it takes for shock formation (ts ). It is instructive to consider a finite-amplitude compression waveform shown in Fig. 11.11 (adopted from Zeldovich and Raizer, 1966) at its initiation at t ¼ to . The waveforms for pressure, acoustic speed, and density are similar in shape. For compression waves, all three non-dimensional parameters—ð p  p0 Þ=p0 , ða  a0 Þ=a0 , and ðρ  ρ0 Þ=ρ0 —are all positive at all times. The increase in ða  a0 Þ=a0 , or simply in a, is the direct result of increase in temperature due to compression. In Fig. 11.11, the similarity variable ð x  ao tÞ is used as the horizontal coordinate, so that the waveform spans over the same wavelength λ at all times. Points on the waves having the same amplitude travel with the same velocity and thus maintain a fixed separation

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11 Unsteady One-Dimensional Flows and Nonlinear Waves

(

p−p0 p−p0 a−a0 or , or p0 a0 p0

)

t0

t1

t2

t3 s

d¢

d

d¢¢

c¢

c¢¢

c

(xs−a0t)

λ

x−a0t

Figure 11.11 Evolution of the shape (or waveform) of a finite-amplitude compression wave. Note that the horizontal axis is similarity variable is ð x  ao tÞ.

between them (such as points c and c’ in on the initial waveform). The finiteamplitude, nonlinear nature of the wave causes the waveform to begin to distort for t > to , as shown in Fig. 11.10. Recall that compression results in increases in temperature and acoustic speed, with larger compression ratios causing larger increases. For example, point d on the initial waveform thus travels faster than, and will at some later time catch up with, point c, causing steepening of the leading edge and flattening of the trailing edge. Prior to shock formation, the steepening (or distortion) of the wave form can be understood as follows. Across the Cþ characteristics generated by the piston shown in Fig. 11.10, J  remains constant: J ¼ u 

2a 2ao ¼ constant ¼  γ1 γ1

which in turn gives u¼

2 ða  a o Þ γ1

The slope of a Cþ characteristic is

dx γþ1 2ao a ¼uþa¼ dt Cþ γ1 γ1

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11.9 Compression Wave and Shock Formation

311

The speed of leading edge of the right-running nonlinear wave has an increasing ð p  p0 Þ=p0 and ða  a0 Þ=a0 . Therefore ðdx=dtÞCþ increases monotonically with a. As a result the later characteristics/waves catch up with the earlier ones, causing the steepening of the gradient and the increasingly distorted waveform, as shown in Fig. 11.11; for example, point d is catching up with point c. On the other hand, the trailing edge becomes increasingly flattened, resulting from the decreasing ð p  p0 Þ=p0 and ða  a0 Þ=a0 . On the ð x  ao tÞ coordinate, the trailing edge lags and flattens behind the leading edge while the leading steepens. At t ¼ t2 , a portion of the leading front in the neighborhood of point s becomes vertical with points c and d being at the same location, xs , and infinite gradients in pressure and density. In Fig. 11.11, this time is designated as t2 , and it is the time of formation of the first shock. At t ¼ t3 , point d moves ahead of point c and the waveform becomes triplevalued for a given ð x  ao tÞ. The multivalued waveform is mathematically impossible; it is also physically absurd in that in a one-dimensional flow no physical mechanism exists for such overtaking (note that in multidimensions, overtaking can occur in rotational flows) and some physical mechanism must be formed at t ¼ t2 (t1 < t2 < t3 ) to prevent this from occurring. It is physically plausible for two values to exist within a few mean free paths (Vincenti and Kruger, 1975). Mathematically, one then obtains an infinitely steepened wave that is a shock wave (Zeldovich and Raizer, 1966), with an infinite gradient at point s in Fig. 11.11. It is said that at t2 , the first shock already forms. The impossible waveform for t ¼ t3 is shown by the dashed line in Fig. 11.11. The location of point s in Fig. 11.11 can be expressed in terms of the similarity variable (xs  a0 ts ), which in turn requires specifying both xs and ts . The following discussion is based on Thompson [1972]. To express xs in terms of a0 and Xp ðtÞ, consider this location in a t-x plot shown in Fig. 11.12. Assume that the two interþ to form the first shock emanate from the piston secting characteristics Ciþ and Ciþ1 path, at two arbitrary times tand t þ dt with corresponding piston locations of xp;i and xp;i þ dx, respectively. The two characteristics intersect after a time period of ts  t at þ characteristic can be imagined to have emanated at the same time as the xs . The Ciþ1 þ Ci characteristic. Then its point of origination will be at (xp;i  dl). Geometrical consideration of the diagram of Fig. 11.12 leads to ts  t ¼

dl dðu þ aÞ

(11.69)

þ That is the extra distance, dl, that the Ciþ1 characteristic has to travel to meet the Ciþ characteristic, due to the change in the slope of the right-running characteristic over the time period from t and t þ dt. Once dl is found, then ts and xs can be determined. Before the intersection occurs, J  remains constant through the flow field. Thus

a ¼ ao þ

γ1 u 2

Applying this result to the slope of the Cþ characteristic leads to

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312

11 Unsteady One-Dimensional Flows and Nonlinear Waves

. up (t) = Xp (t)

t Xp (t)

( dxdt )

C+i+1

ts

= (u+a) + d (u+a)

C+i+1 Ci+ dx = (u+a) dt C+i

C+

dt

ts–t

( )

t dl

dx

ts

L xp,i–1

xp,i

xs

xp,i+1

x

Figure 11.12 The t-x plot showing the time (ts ) and location (xs ) where the shock wave forms due to a piston moving with a speed up ðtÞ, increasing with time, into an initially quiescent gas in a long tube.



dx dt

Cþ

¼ u þ a ¼ ao þ

γþ1 γþ1 γ þ 1 dXp u ¼ ao þ up ¼ a o þ 2 2 2 dt

(11.70)

Therefore, dðu þ aÞ ¼

γ þ 1 d2 Xp dt 2 dt2

Considering a general Cþ characteristic in Fig. 11.12, one arrives at



dx γ þ 1 dXp dl þ dx ¼ dt ¼ ao þ dt dt Cþ 2 dt   With dx ¼ dXp =dt dt

γ  1 dXp dl ¼ ao þ dt 2 dt

(11.71)

(11.72)

Substituting Eqns. (11.70) and (11.72) into Eqn. (11.69) yields

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11.9 Compression Wave and Shock Formation


 i dXp

ao þ γ1 2 dt dt 1 2 γ1 _
 2 ao þ ¼tþ ts ¼ t þ Xp €p γ þ 1 γþ1 d Xp γþ1 X dt 2 dt2

313

h

(11.73)

  € p ¼ d2 Xp =dt2 . where X_ p ¼ dXp =dt ¼ up and X Note that Eqn. (11.73) is derived for an arbitrary time when two characteristics intersect. For the earliest moment of shock formation, ts takes up a minimum value ts;min :



 2  1 γ  1 dts 2 γ1 _ € € € ⃛ ao þ ¼ 0 ¼ 1  Xp Xp Xp þ Xp Xp γþ1 γþ1 γþ1 dt The above expression leads to an implicit equation for ts;min :

 2 1 γ1 _ €p ¼ ⃛p ao þ Xp X X γ 2

(11.74)

The corresponding location where the first shock forms is



  γþ1 _ 1 2 γ1 _ a þ xs;min ¼ Xp þ ðu þ aÞ ts;min  t ¼ Xp þ ao þ Xp X €p γ þ 1 o γ þ 1 p 2 X (11.75) The traveling shock formed by intersection of Cþ characteristics, such as those shown in Figs. 11.9 and 11.10 are call the right-traveling or Cþ shocks. The flow field after shock formation clearly has become piecewise homentropic, as entropy is only constant (equal to different values) in uniform regions. In Fig. 11.9, regions I and II are both homentropic, with region II having higher entropy than region I. In Fig. 11.10, the undisturbed region I (shaded) is a homentropic region, while in region II (non-shaded), only that part of the t-x plane bounded by the piston path line and the limiting C characteristic is homentropic. Consider Xp ¼ Að1  cos ωtÞ in a long tube containing quiescent gas at t ¼ 0. Assume that A ¼ a1 =2ω. Find the time and the location of the first shock formation.

EXAMPLE 11.7

Solutions – Equation (11.74) provides

  1 γ1 ao þ Aω sin ωt Aω3 sin ωt A ω cos ωt ¼ γ 2   Substituting A ¼ ao =2ω and cos 2 ωt ¼ 1  sin 2 ωt into the above expression and rearranging leads to 2

4

2

 ð γ þ 1Þ 1 1 sin 2 ωt þ sin ωt þ ¼ 0 8γ 2γ 4

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314

11 Unsteady One-Dimensional Flows and Nonlinear Waves

The solution for this quadratic equation for γ ¼ 7=5 (for diatomic gases) is
pffiffiffiffiffi sin ωt ¼ 5  67 =6 Because j sin ωtj ≤ 1, the above solutions yield only meaningful results of sin ωt ≈ 0:531 ts;min ≈

0:560 ω

If ω ¼ 1 s1 , then ts;min ¼ 0:56 s. The value of ts;min depends on both the amplitude, A, and ω. In this example, A ¼ a1 =2ω is a function of ω, leaving ts;min solely dependent on ω. The value of A is chosen so that X_ p ¼ ða1 =2Þ sin ωt, ensuring that the amplitude of the piston velocity is not too small compared to the acoustic speed in the quiescent gas for the nonlinear theory to apply. The corresponding location where the first shock forms is

  γþ1 _ Xp xs;min ¼ Xp þ ðu þ aÞ ts;min  t ¼ Xp þ a1 þ 2

1 2 γ1 _ a1 þ Xp € γþ1 Xp γ þ 1 With Xp ¼ Að1  cos ωtÞ and A ¼ a1 =2ω andγ ¼ 7=5 xs;min ≈ 2:881

a1 ω

□ 11.10 Flow with Weak Traveling Shock The relationship between jumps in entropy and pressure across a weak shock wave has been shown in Chapter 4 as  2 3 γ þ 1 3 s2  s1 2γ ≈ M1  1 ¼ β 2 12γ2 R 3ð γ þ 1Þ

(4.20f)

where the shock strength is given by Eqn. (4.18d): β≡

 p2  p1 2γ  2 M1  1 ¼ γþ1 p1

(4.16e)

For a weak shock, β ≪ 1. The entropy change across a weak shock wave is of the order of β3 , which is negligibly small. Therefore, the flow field with weak shocks can be approximated as homentropic, and the above results from the method of characteristics that can be used for flow calculations, with negligible changes in the Riemann invariants. Figure 11.13 schematically shows the invariants and the particle

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11.10 Flow with Weak Traveling Shock (a)

315

(b) t C– Characteristics (J–2 = J–1) Particle path

C– shock Particle path

C+ shock

C+ Characteristics (J+2 = J+1) (2) (2)

(1)

(1)

x

Figure 11.13 Schematics showing the characteristics and associated invariants and the particle trajectory across a Cþ and C shocks traveling into a quiescent gas.

trajectory across a Cþ and a C shocks traveling into a quiescent gas, respectively. The following derivation shows the negligible changes in the invariants across a weak shock, which in turns requires showing negligible changes in both the flow velocity, u, and the acoustic speed, a. Take a Cþ shock as an example. The change in u across the weak Cþ shock can be obtained by requiring that J2 ¼ J1 along a C characteristic, shown in Fig. 11.13a. Therefore, by combining the result of Problem 11.5, pð2

u2  p0



pð1



dp dp ¼ u1  ρa s ρa s p0

and pð2

u2  u1 ¼ p1



dp ρa s

(11.76)

The following expression is useful for the integration to be carried out for Eqn. (11.76)



∂–v d– v ∂ρ 1 ¼ ¼ 2 2 (11.77) ∂p s dρ ∂p s ρa For small increases in p from p1 , a Taylor series in ð p  p1 Þ is useful by considering the state ahead of the shock wave (state 1 in Fig. 11.13) as the reference state and expanding:



2



∂– v ∂– v ∂ – v 1 ∂3 –v ¼ þ ð p  p1 Þ þ ð p  p1 Þ2 þ … (11.78) ∂p s ∂p s;1 ∂p2 s;1 2 ∂p3 s;1 By repeating the similar steps shown in Eqn. (11.77),

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316

11 Unsteady One-Dimensional Flows and Nonlinear Waves



∂2 – v ∂p2









d ∂–v ∂ρ d 1 ∂ρ 2  2 2 ¼ ¼ dρ ∂p ∂p s;1 dρ ρ a ∂p s;1 ρ31 a41

¼ s;1

and

∂3 – v ∂p3



¼ s;1







d ∂2 – v ∂ρ d 2 ∂ρ 6 ¼ ¼ 4 6 3 4 2 dρ ∂p ∂p s;1 dρ ρ1 a1 ∂p s;1 ρ1 a1

With these two expressions, Eqn. (11.78) becomes

∂– v 1 2 3 ¼  2 2 þ 3 4 ð p  p1 Þ  4 6 ð p  p1 Þ 2 þ … ∂p s ρ1 a1 ρ1 a1 ρ1 a1

(11.79)

Substituting Eqn. (11.77) into Eqn. (11.76) yields pð2

u2  u1 ¼ p1

pð2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi



dp ∂–v  dp ¼ ρa s ∂p s p1

Using Eqn. (11.79), one obtains 1 u 2  u1 ¼ ρ1 a1

pð2

p1

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   2 3 2 ð p  p Þ  ð p  p Þ 1 dp 1 1 ρ1 a21 ρ21 a41

Performing a binomial expansion, for small ð p  p1 Þ, and integration for this expression leads to   1 1 1 1 1 2 3 u 2  u1 ¼ ð p 2  p1 Þ  ð p  p Þ þ ð p  p Þ (11.80) 2 1 2 1 ρ1 a1 2 ρ1 a21 2 ρ21 a41 The relative change in gas velocity compared to the acoustic velocity in state 1 is, after using β ¼ ðp2  p1 Þ=p1 ¼ γðp2  p1 Þ=ρ1 γRT1 ¼ γðp2  p1 Þ=ρ1 a21 , u2  u1 1 1 1 ¼ β  2 β2 þ 3 β3 ðfor a Cþ shock with β ≪ 1Þ γ 2γ 2γ a1

(11.81)

For comparison, the velocity change across a stationary normal shock wave is, following Eqns. (4.20c) and (4.20d),  V2  V1 2M1  2 β M1  1 ≈  ðfor a stationary normal shock with M1 ! 1 ≈ γ a1 γþ1 and β ≪ 1Þ

(11.82)

The difference of velocity change across a weak stationary and a weak moving shock is of O(β2 Þ, which is even smaller than the pressure change. It is noted that the signs of the leading terms are opposite, for the following reasons. Moving shock waves carry gas along with them and thus increase the gas velocity (from zero). Across

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11.10 Flow with Weak Traveling Shock

317

a stationary shock, the increase in pressure is accompanied by a decrease in gas velocity. The change in acoustic speed across the Cþ shock is related to the change in static temperature. For both a stationary or a moving shock  T2 2γðγ  1Þ  2 ≈1 þ M1  1 2 T1 ð γ þ 1Þ

(4.22b)

Therefore, a2  a1 ¼ a1

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi  T2 2γðγ  1Þ  2 M1  1  1 1¼ 1þ 2 T1 ðγ þ 1Þ

By using Eqn. (4.16e) and binomial expansion for small β, the above expression becomes a2  a 1 γ1 β ¼ 2ð γ þ 1Þ a1

(11.83)

Subtracting Eqn. (11.83) from Eqn. (11.81), the change in J  across the weak Cþ shock is J2  J1 1 1 1 β  2 β2 þ 3 β 3 ¼ γðγ þ 1Þ 2γ 2γ a1

(11.84)

which is ≪ 1,validating the requirement of J2 ¼ J1 and allowing applicability of the method of characteristics for weak shocks. A long piston-cylinder device contains air having a temperature of 300 K (a1 ¼ 347:8 m=s) and a pressure of 1 atmosphere. The air is initially at rest and then the piston is impulsively moved to compress the air at a speed of 100 m/s. Find the speed of the resultant shock before it reaches the end of the cylinder and the associated pressure and temperature increases. EXAMPLE 11.8

Solutions – Consider the shock to be a right-running (Cþ ) shock. Although ðu2  u1 Þ=a1 ≈ 0:29 is not much smaller than unity, one might make an initial attempt to solve the problem using the method of characteristics. In the simple wave region, the method of characteristics requires J1 ¼ J2 . Therefore, u2 

a2 ¼ a1 þ

2 2 a2 ¼  a1 γ1 γ1

γ1 u2 ¼ 347:8 m=s þ 0:2  100 m=s ¼ 367:8 m=s 2

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318

11 Unsteady One-Dimensional Flows and Nonlinear Waves

With the weak shock assumption, it is reasonable to calculate the wave speed, Vs , as  

1 dx dx 1 Vs ¼ þ ¼ ½ða1 þ u1 Þ þ ða2 þ u2 Þ ¼ 407:8 m=s 2 dt 1 dt 2 2 With the isentropic assumption for the method of characteristics, p2 ¼ p1

γ=ðγ1Þ 2γ=ðγ1Þ T2 a2 ¼ ¼ 1:48 T1 a1 T2 ¼ T1

2 a2 ¼ 1:12 a1

Comments – (1) For a traveling shock with a speed of 407:8 m=s into the quiescent air, the Mach number is calculated using Eqn. (4.22d):  V1  V2  V2 2  2 2  2 M1  1 or M1  1 ≈1  ≈ γþ1 γþ1 V1 V1 For an observer riding on the V2 ¼ Vs  u2 ¼ 307:8 m=s. Therefore,

wave,

V1 ¼ Vs ¼ 407:8 m=s

and

Ms ¼ M1 ¼ 1:16 for which, p2 ¼ 1:40 p1 T2 ¼ 1:10 T1 Therefore, solutions using the method of characteristics and shock relationship yield very similar results, although both β ≈ 0:4 and X_ p =a1 ¼ u2 =a1 ≈ 0:29 are not much smaller than unity. (2) If the shock is left-running (C ), one would obtain the same results – this is □ left as an exercise problem.

11.11 Wave Interaction A non-simple wave region is where waves interact. Example 11.5 illustrates the interaction between compression and expansion waves. There are other types of interaction. They generally fall into two categories. The first category is that of

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11.11 Wave Interaction

319

up

Particle path

t

Uniform region 3 u=0

Simple region

. up = Xp (t)

9 7 4

Uniform region 2 u = up

3

6 2

10 Non-simple 8 region; (wave 5 interaction region) 1

Simple region Uniform region 1 u=0

Figure 11.14 Characteristics and particle path resulting from a piston moving with a constant speed up in a long tube away from the gas bounded by the piston and the closed end of a long tube.

characteristics (i.e., isentropic waves and weak shocks) and the second of shock waves. Consider the situation where centered expansion waves are generated by a piston in a long tube with a rigid end wall, as shown in Fig. 11.14. The rightrunning rarefaction, or expansion, wave (Cþ characteristics) will be turned back by the wall to form left-running rarefaction waves (C characteristics). These Cþ and C characteristics interact in the near-wall region to create the non-simple region, shown as the shaded region in Fig. 11.14. The regions outside the simple and nonsimple regions are uniform regions such as regions 1, 2, and 3. In both uniform regions 1 and 3, the gas velocity is zero, while in uniform region 2 the gas velocity is equal to that of the piston impulsively set in motion with a constant velocity up ¼ X_ p , at t ¼ 0. Gas velocities in these uniform regions are required to satisfy the stationary and moving boundaries of the end wall and the moving piston, respectively. Two representative particle trajectories are given in Fig. 11.14, with one going through only the simple wave and uniform regions and the other through both simple- and non-simple wave regions. It is noted that an expansion wave pushes particles in the direction opposite to its propagation, as qualitatively described by these two trajectories.

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320

11 Unsteady One-Dimensional Flows and Nonlinear Waves (a)

Particle path

(b)

C− shock 3

C+ shock

3

2

2 1

C+ shock

1 C− shock

x, ub

x, ub

Figure 11.15 Incident and reflected shock waves from a wall that is either stationary (ub ¼ 0) or moving (ub ≠ 0 ); Jiþ ¼ Jr at a stationary wall, and Jrþ ¼ Ji þ 2ub at a moving wall.

Jiþ

In Fig. 11.15a the incident Cþ and reflected C shocks carry invariant values of and Jr , respectively: ð dp þ (11.85) Ji ¼ u þ b ρa Jr

ð ¼u

dp b ρa

(11.86)

where subscripts i and r denote incident and reflected waves, respectively. Recall that in general the integration in Eqns. (11.85) and (11.86) results in different values from one to another characteristic. However, since both the Cþ and C characteristics share the common wall condition, the integration represent a thermodynamic condition at the wall (designated by the subscript, b). At the stationary wall surface, u ¼ 0, and thus for a pair of incident and reflected characteristics at the wall Jiþ ¼ Jr ðat a stationary wallÞ

(11.87)

In general, the wall boundary might be moving with a velocity u ¼ ub , then Jr ¼ Jiþ þ 2ub ðat a moving wallÞ

(11.88)

If the reflection is from a left boundary, then the incident wave is a C characteristic (Fig. 11.15b) and Jrþ ¼ Ji þ 2ub It is seen in Fig. 11.15 that whether or not ub ¼ 0, the particle trajectory at the boundary is that of the moving wall as required by the boundary conditions. Due to thin shock waves, regions 1, 2, and 3 in Fig. 11.15 are each a uniform region. The wall boundary in Fig. 11.15a is moving with a constant speed ub ¼ a1 =8 to the right while the gas behind the incident wave is traveling at

EXAMPLE 11.9

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11.11 Wave Interaction

321

a speed equal to a1 =4. Find the pressure in each of regions 1, 2, and 3, given that the pressure in region 1 is p1 . Also find the pressures for ub ¼ 0 Solution – u1 ¼ ub ¼ a1 =8 u2 ¼ a1 =4 u3 ¼ ub ¼ a1 =8 In the simple wave region across the Cþ characteristic 2a1 2a2 1 2a1 1 2a2 ¼ u2  ; that is a1  ¼ a1  8 γ1 γ1 γ1 4 γ1

J  ¼ u1 

Across the C characteristic J þ ¼ u2 þ

2a2 2a3 1 2a2 1 2a3 ¼ u3 þ ; that is a1 þ ¼ a1 þ 4 γ1 γ1 γ1 8 γ1

Therefore, a2 ¼ a1 þ

γ1 a1 16

a3 ¼ a1 þ

γ1 a1 8

Applying isentropic relations for MOC yields: p2 ¼ p1 p3 ¼ p2



T2 T1

γ=ðγ1Þ

2γ=ðγ1Þ a2 γ þ 15 2γ=ðγ1Þ ¼ ¼ 16 a1

 γ=ðγ1Þ 2γ=ðγ1Þ  T3 a3 2ðγ þ 7Þ 2γ=ðγ1Þ ¼ ¼ γ þ 15 T2 a2

For γ ¼ 1:4, p2 =p1 ¼ 1:19 and p3 =p2 ¼ 1:18. For ub ¼ 0, J  ¼ u1 

2a1 2a2 2a1 1 2a2 γ1 a1 ¼ u2  ;  ¼ a1  and a2 ¼ a1 þ 8 γ1 γ1 γ1 4 γ1

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322

11 Unsteady One-Dimensional Flows and Nonlinear Waves

J þ ¼ u2 þ

2a2 2a3 1 2a2 2a3 γ1 a1 ¼ u3 þ ; a1 þ ¼ and a3 ¼ a1 þ 4 γ1 γ1 4 γ1 γ1 p2 ¼ p1 p3 ¼ p2

2γ=ðγ1Þ a2 γ þ 7 2γ=ðγ1Þ ¼ 8 a1

 2γ=ðγ1Þ  a3 2ðγ þ 3Þ 2γ=ðγ1Þ ¼ γþ7 a2

For γ ¼ 1:4, p2 =p1 ¼ 1:41 and p3 =p2 ¼ 1:38. Comments – Because the gas is traveling at a higher speed (a0 =4) after the wave passes, one expects the wave to be a shock wave. Therefore, it is instructive to compare the above results with shock solutions. For simplicity, consider the case of ub ¼ 0; this is left as an end-of-chapter problem, where one result is that ub ¼ 0 leads to □ a harder hit of the wave on the wall than that in this example. Consider a long tube, as described in Fig. 11.14, with the right end closed and the left end being a movable piston. The gas in the tube is air and is initially at rest. A centered expansion is then generated by withdrawing the piston impulsively to a velocity up . Find the flow conditions in the non-simple wave region.

EXAMPLE 11.10

Solutions – For convenience, choose characteristics in the expansion fan that have velocities equal to 0, up =3, 2up =3, and up . The non-simple wave region is shown by the shaded area in Fig. 11.14. A table similar to that of Example 11.5 helps to summarize the calculation results, as shown below, where the subscript “o” designates the undisturbed region: Comments – 2a1 2a2 ¼ u2  γ1 . Knowing For point 2, J  from point 1 can be used so that u1  γ1 u1 ¼ 0 u2 ¼ up =3 and a1 ¼ a0 , the same results for a2 and u2 are found. 2a3 2a2 ¼ u3  γ1 Similarly, for point 3, J  from point 2 can be used so that u2  γ1 leading to the same results for a3 and u3 as shown in the table. To increase the accuracy of the flow conditions (i.e., u, a, and the slopes of the characteristics, u þ a and u  a), more characteristics emanating from point O can be chosen or the slopes of characteristics can be taken as the average value of those at the two points connected by the given characteristic. For example the slope of the characteristic segment connecting points 6 and 7 is now

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Point

Invariants þ

Found

uþa

uo ¼ 0

a1 ¼ a0

u 1 þ a1 ¼ a0 u 2 þ a2 ¼ a0 

J ¼ uo þ

2

2a0 2a2 J  ¼ u0  γ1 ¼ u2  γ1

u2 ¼ up =3

a2 ¼ a0  γ1 6 up

3

2a0 2a3 J  ¼ u0  γ1 ¼ u3  γ1

u3 ¼  23 up

a3 ¼ a0  γ1 3 up

4

2a0 2a4 J  ¼ u0  γ1 ¼ u4  γ1

u4 ¼ up

a4 ¼ a0  γ1 2 up

5

2a5 2a2 J þ ¼ u2 þ γ1 ¼ u5 þ γ1

u5 ¼ 0

a5 ¼ a0  γ1 3 up

6

J þ ¼ u3 þ 

J ¼ u5  7

þ

J ¼ u4 þ J  ¼ u6 

8

J þ ¼ u6 þ

9

J þ ¼ u7 þ 

J ¼ u8  10

J þ ¼ u9 þ

2a3 γ1 2a5 γ1 2a4 γ1 2a6 γ1 2a6 γ1 2a7 γ1 2a8 γ1 2a9 γ1

¼ u1 þ

Known

1

2ao γ1

¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼

2a1 γ1

2a6 u6 þ γ1 2a6 u6  γ1 2a7 u7 þ γ1 2a7 u7  γ1 2a8 u8 þ γ1 2a9 u9 þ γ1 2a9 u9  γ1 10 u10 þ 2a γ1

u1 ¼ 0




 þ γ1 6 up u3 þ a3 ¼ a0  23 þ γ1 3 up
γ1 u 4 þ a4 ¼ a0  1 þ 2 u p 1
3

¼ u1  a1 ¼ a0

  γ1 6 up u3  a3 ¼ a0  23  γ1 3 up
γ1 u4  a4 ¼ a0  1  2 up u2  a2 ¼ a0 




1
3

u6 ¼  13 up u7 ¼  23 up

u 7 þ a7 ¼ a0 

Þ a8 ¼ a0  2ðγ1 3 up

u9 ¼  13 up

Þ u8 þ a8 ¼ a0  2ðγ1 up
3  5ðγ1Þ 1 u 9 þ a9 ¼ a0  3 þ 6 up

Þ u8  a8 ¼ a0 þ 2ðγ1 up
3  5ðγ1Þ 1 u9  a9 ¼ a0  3  6 up

a10 ¼ a0  ðγ  1Þup

u10 þ a10 ¼ a0  ðγ  1Þup

u10  a10 ¼ a0 þ ðγ  1Þup

Þ a7 ¼ a0  2ðγ1 3 up

Þ a9 ¼ a0  5ðγ1 6 up

u10 ¼ 0

dt 12

u5 þ a5 ¼ a0  γ1 up
3  u6 þ a6 ¼ a0  13 þ γ1 2 up

a6 ¼ a0  γ1 2 up

u8 ¼ 0

ua dx




2 3

 Þ þ 2ðγ1 up 3

u5  a5 ¼ a0 þ γ1 up
3  u6  a6 ¼ a0  13  γ1 2 up u7  a7 ¼ a0 




2 3

 Þ  2ðγ1 up 3

324

11 Unsteady One-Dimensional Flows and Nonlinear Waves

 ðu  aÞ ¼ a0 

 1 7ð γ  1Þ  up : 2 12

Once the slopes are determined, they are used to determine the locations of the intersections of these characteristics. □

11.12 The Shock Tube The capability of achieving rapid and steep temperature increases finds important applications in gas-phase chemical kinetics – for example, determining the temperature for ignition and the ignition delay time under prescribed pressures. Rapid temperature rises can be generated by passing a shock wave through the gas medium. In practice, the shock tube provides a means for such a purpose. The schematic of a simple shock tube of finite length is shown in Fig. 11.16a. The diaphragm separates the low-pressure (driven) gas from the high-pressure (driver) gas. For chemical kinetics and ignition studies, the low-pressure, driven gas is the gas to be tested (the test gas that is a reactive mixture). For convenience, the pressure-temperature pairs of the driving and driven gases are ðp4 ; T4 Þ and ðp1 ; T1 Þ, respectively. The diaphragm only has a limited mechanical strength designed to rupture, resulting from a sufficient pressure difference across it or by other mechanical and/or electrical means. The initial pressure ratio (p4 =p1 ) sets up the condition for the shock strength and the associated temperature rise. The diaphragm would ideally disappear upon its rupture, to avoid flow tripping that causes the flow to cease to be one-dimensional. After the rupture takes place at t ¼ 0, the following observations are made. (1) The pressure discontinuity propagates in two directions: as a shock into the driven/test gas 1, and as a centered rarefaction/expansion wave into the driving gas. A snapshot of the flow regimes at a particular (though arbitrary) time t ¼ t1 , prior to occurrences of wave reflection from end walls, is shown in Fig. 11.16b. Figures 11.16c and 11.16d show for t ¼ t1 the pressure and temperature distributions, respectively, throughout the shock tube. At t ¼ t1 , the flow regimes within the shock tube are either of the uniform or the simple-wave types, as shown in the wave diagram shown in Fig. 11.16e. The wave pattern as time proceeds, for example, t4 > t3 > t2 > t1 , is also illustrated in Fig. 11.16e. (2) The contact surface (shown in Fig. 11.16b), which is a materials surface and behaves like a gas particle, follows the shock wave in the direction toward the driven gas, as a compression wave carries gas particles. Reasoning based on the expansion wave propagation leads to the same result, as the left-propagating expansion wave pushes gas and the contact surface in the opposite direction, that is, toward the driven gas. In region 2, the particle path, represented by XPP ðtÞ, in Fig. 11.16e, is parallel to that of the contact surface, just like the gas near the surface of a moving piston.

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11.12 The Shock Tube

(a)

325

t=0 Low-pressure gas p1, T1

High-pressure gas p4, T4

Diaphragm (initial contact surface) (b)

(c)

t = t1 4

3

2

1

V4 = 0

V3 = V2

V2

VS = WIS

Contact surface

t = t1

Normal shock

p4 p3 = p2

p1

(d)

t = t1 T2 (>T1)

T4(=T1)

T1 T3 < T4

Figure 11.16 The shock tube of finite length – (a) initial state (e.g., before the diaphragm is ruptured); (b) the waves and the states of gases at t ¼ t1 shortly after the rupture; (c) pressure distribution at t ¼ t1 ; (d) temperature distribution at t ¼ t1 ; (e) characteristics and wave patterns in the t-x plot.

(3) At t ¼ t1 , pressures and velocities across the contact surface match, so that p2 ¼ p3 (Fig. 11.6c) and u2 ¼ u3 . However, in general T2 ≠ T3 (Fig. 11.16d), as T2 is raised from T1 by shock and T3 decreases from T4 by expansion. (4) The pressure ratio p2 =p1 determines the Mach number of the shock wave and the temperature ratio T2 =T1 . However, p2 =p1 and, consequently, T2 =T1 are related to the initial pressure ratio p4 =p1 . (5) At a later time, expansion and the shock waves reflect from the end walls. For example, the incident shock reflects from the wall at t ¼ t2 (see Fig. 11.16e). The reflected shock meets the contact surface at point a at t ¼ t3 . Since p2 ¼ p3 and u2 ¼ u3 , the reflected shock would propagate into region 3 without

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326

11 Unsteady One-Dimensional Flows and Nonlinear Waves t

(e)

xpp(t)

6 7

t = t4

a 5

su

rfa

ce

3 Expansion fan

2

τmax

Co n

ta

ct

t = t2 t = t1

ck

4

t = t3

Sho

1

O xpp(t = 0)

x

Location of observation

Figure 11.16 (cont.)

interference. In practice, reflection may occur at the contact surface due to several factors, among which are the acoustic impedance resulting from the driver, the driven gases being different, and the non-ideal conditions such as boundary layer development in the shock-induced flow. In Fig. 11.16e, the reflected shock by the contact surface is shown, separating regions 5 and 6. In region 5, u5 ¼ 0 and the gas particle is motionless as depicted by the XPP ðtÞ in the wave diagram. At t ¼ t2 the shock wave is reflected from the end wall to propagate into region 2. The reflected shock meets the contact surface at t ¼ t3 (i.e., point a, as shown in Fig. 11.16e). For chemical kinetics studies, T2 is the temperature of interest as well as the time (between the passing of incident shock and t3 ) the test gas spends at this temperature. To maximize the test time, the location for observation is chosen for the duration τmax shown in Fig. 11.16e. It is of interest to find T2 =T1 and τmax as related to the initial pressure ratio, p4 =p1 , as follows. Assume the gases (or gas mixtures) in Fig. 11.16a are ideal gases. Across the leftrunning expansion wave (i.e., along Cþ characteristics: also refer to Fig. 11.16b or 11.16e for subscripts), J þ ¼ u3 þ

2a3 2a4 ¼ u4 þ γ1 γ1

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11.12 The Shock Tube

327

For expansion, isentropic relation applies. By rearranging the above expression, the pressure ratio across the expansion wave is p3 ¼ p4



T3 T4

γ4 =ðγ4 1Þ

2γ4 =ðγ4 1Þ

a3 γ4  1 u3 2γ4 =ðγ4 1Þ ¼ ¼ 1 2 a4 a4

(11.89)

where γ3 ¼ γ4 and u4 ¼ 0. This expression suggests that p3 and therefore p2 (¼ p3 ) are determined once u3 is known. Because u3 ¼ u2 (at the contact surface), the shock relations can be used to find u3 . By combining the following equations β≡

 p2  p1 2γ1  2 M1s  1 ¼ p1 γ1 þ 1

(4.16e) or (11.9)

with M1s being the Mach number with which the shock propagates into the quiescent gas region 1 (u1 ¼ 0). From Eqn. (4.20d) for stationary shock:

u 2  u1 1 β 2 1 M1  ¼ ¼ M1 ρ1 a21 =p1 γ1 þ 1 M1 a1 one finds u2 ¼



2 1 a1 M1s  ¼ u3 γ1 þ 1 M1s

for the moving shock. Equation (11.17) can be used for M2 :

rffiffiffiffiffiffi 2 1 T1 M1s  M2 ¼ γ1 þ 1 M1s T2

(11.90a)

(11.90b)

Combining Eqns. (11.89) and (11.90) yields 

2γ4 =ðγ4 1Þ p3 γ4  1 a1 1 ¼ 1 M1s  γ1 þ 1 a4 M1s p4

(11.91)

The propagation Mach number M1s is found by again using Eqn. (4.16b): 2  ðγ1  1Þ p2 2γ1 M1s ¼ γ1 þ 1 p1

(11.92a)



γ1 1 2γ1 ðγ1 1Þ=γ1 2 2 1 þ M M  1 1s 1s γ 1 2 T2 p2 h i1 ¼ ¼ ðγ1 þ1Þ2 T1 p1 2 M 2ðγ1 1Þ

(11.92b)

1s

The matching condition p2 ¼ p3 leads to M1s as a function of the initial pressure ratio p4 =p1 , as shown in the following implicit equation for M1s . 

2γ4 =ðγ4 1Þ   2 2γ1 M1s  ðγ1  1Þ p4 p4 p3 p2 γ4  1 a1 1 ¼ ¼ 1 M1s  (11.93) γ1 þ 1 a4 M1s γ1 þ 1 p1 p3 p2 p1

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11 Unsteady One-Dimensional Flows and Nonlinear Waves

For strong shock applications (for large T2 =T1 , to initiate chemical reaction), p4 =p1 ! ∞, which with γ4 =ðγ4  1Þ > 0 and the second bracket is > 0, requiring the expression in the first bracket in Eqn. (11.93) to approach zero. Therefore, M1s 

1 γ þ 1 a4 ! 1 M1s γ4  1 a1

(11.94)

For strong shock application, M1s ≫ 1, yielding M1s ≈

γ1 þ 1 a4 γ4  1 a1

(11.95)

It is desirable to complete the flow conditions at t ¼ t1 and those in region 5. By using pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 ¼ u3 , M3 γ3 R3 T3 ¼ M2 γ2 R2 T2 . For ideal gases γ2 ¼ γ1 , γ3 ¼ γ4 , R2 ¼ R1 and R3 ¼ R4 . Therefore, the velocity matching at the contact surface becomes pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M3 γ4 R4 T3 ¼ M2 γ1 R1 T2 Across the simple wave regime, J þ ¼ u þ 2a=ðγ  1Þ ¼ constant. With u4 ¼ 0, 2 2 2 a4 ¼ u3 þ a3 ¼ u 2 þ a3 γ4  1 γ4  1 γ4  1 Rearranging the above expression while using Eqn. (11.90), one obtains

γ4  1 1 a4 ¼ a3 þ a1 M1s  (11.96) γ1 þ 1 M1s pffiffiffiffiffiffiffiffiffi With a ≡ γRT , Eqn. (11.96) can be rewritten to obtain the temperature in region 3 as "

#2

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi T3 γ4  1 γ1 R1 T1 1 ¼ 1 M1s  γ1 þ 1 M1s T4 γ4 R4 T4

(11.97)

The pressure in region 3 is given by 2  ðγ1  1Þ p3 p2 2γ1 M1s ¼ ¼ γ1 þ 1 p1 p1

(11.98)

It is noted that M1s is found is using the implicit equation, Eqn. (11.93). Assume that the test gas in a shock tube experiment consists of a very dilute fuel-air mixture, the driving gas is nitrogen, and that the initial temperature is uniform throughout the entire tube. Find for the strong shock condition, the initial pressure ratio required, and the temperature and the pressure under which the experiment is being conducted.

EXAMPLE 11.11

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11.12 The Shock Tube

329

Solution – For both nitrogen and the dilute mixture, γ1 ¼ γ4 ¼ 1:4 and Eqns. (11.94) and (11.95) yield M1s ¼ 6:16 and 6, respectively. For convenience, take M1s ¼ 6. Thus 2  ðγ1  1Þ p2 2γ1 M1s ¼ 41:8 ¼ γ1 þ 1 p1

The shock relationship gives (for γ ¼ 1:4; or from the shock table) 

γ1 2γ 2 2 1 þ M M  1 1s 1s 2 γ1 T2 h i ¼ ¼ 7:94 ðγþ1Þ2 T1 M2 2ðγ1Þ

1s

To generate the above temperature rise, the required initial pressure ratio is 

2γ4 =ðγ4 1Þ   2 2γ1 M1s  ðγ1  1Þ p4 γ  1 a1 1 ¼ 1 4 M1s  ¼ 3:3  1012 γ1 þ 1 a4 M1s γ1 þ 1 p1 Comments – The initial pressure for the given conditions is too enormous to be realistic. Equation (11.95) suggests that it is desirable to have a combination of (1) a large value of a4 =a1 and (2) a small specific ratio γ4 (i.e., approaching 1). For example, raising T4 helps to simultaneously increase a4 and γ4 . Using polyatomic gas as the driver helps to decrease γ4 . For example, one may use propane (C3H8) at 290 K as the driver gas, so that γ4 ¼ 1:13. Assume also T1 ¼ 290 K. Thus p2 =p1 ¼ 209:3 and T2 =T1 ¼ 4:6. The initial pressure ratio is then p4 =p1 ¼ 4:5  105 , which is more manageable, especially for small p1 . When a desired value of T2 =T1 is known, Eqns. (11.92) and (11.93) are then used to □ find M1s p4 =p1 , respectively. The result of Example 11.11 suggests that to lessen the requirement for large p4 =p1 , observations can be made in region 5 in the wave diagram, because T5 > T2 . Region 5 offers the added benefit that observation is made of the same motionless gas particle. Further benefits can be derived if the acoustic impedances are matched across the contact interface – under this condition there is no reflected shock from the contact and the duration of the observation time is until the reflected expansion wave reaches region 5. It is then of interest to find T5 and p5 as functions of p4 =p1 . First of the velocity and Mach number of the reflected shock should be found, as described in the following. Let the velocity of the reflected shock be Vsr in the x direction, with the gas velocity in region 2 relative to the reflected shock equal to ðVsr þ u2 Þ, while the gas velocity downstream of and relative to the reflected shock (i.e., region 5) is Vsr . Then

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330

11 Unsteady One-Dimensional Flows and Nonlinear Waves

Msr ¼

Vsr þ u2 a2

(11.99)

Equation (11.92a) can be rewritten with M1s replaced by Msr and subscripts 2 and 1 by 5 and 2, respectively. Further assuming that no reaction has been initiated yet, so that γ2 ¼ γ1 , one arrives at

γ þ 1 p5 γ 1 2 Msr ¼ 1 (11.100) þ 1 2γ1 2γ1 p2 Rewriting Eqn. (4.18) with the aid of Eqns. (11.9) and (11.100) for the reflected shock and assuming that γ5 ¼ γ1 , one obtains the following
 2 ðγ1 þ 1Þ pp52 þ ðγ1  1Þ ρ5 ðγ1 þ 1ÞMsr
 ¼ (11.101) ¼ 2 þ2 ρ2 ðγ1  1ÞMsr ðγ  1Þ p5 þ ðγ þ 1Þ 1

T5 ¼ T2

p2

1





 2γ1 γ1 þ1 þ p5 2 2 1 þ γ1 21 Msr γ1 1 Msr  1 γ1 1 p2 p5

 h i ¼ γ þ1 ðγ1 þ1Þ2 p5 p2 1 þ 1 2 γ1 1 p2 2ðγ 1Þ Msr

(11.102)

1

Vsr þ u2 ¼ Msr ¼ a2

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi

γ1 þ 1 p 5 γ1  1 þ 2γ1 2γ1 p2

(11.103)

Equations (11.101) through (11.103) calls for the knowledge of p5 . The following derivation is needed for this purpose. First, look at the relationship between Msr and M1s . Continuity across the incident shock requires Vsr þ u2 ρ5 ¼ Vsr ρ2 Because Vsr þ u2 ¼ Msr a2 ,



ρ u2 ¼ Msr a2 1  2 ρ5

By using Eqn. (11.101) and combining Eqn. (11.90a), one finds



2 1 2 1 u2 ¼ a2 Msr  a1 M1s  ¼ γ1 þ 1 Msr γ1 þ 1 M1s

(11.104)

By using the Rankine-Hugoniot relationship, Eqn. (4.18b), the relation between a1 and a2 can be written as 1 2  

a2 T2 p2 ρ1 p2 γ1 þ 1 p2 γ1 þ 1 p2 þ ¼ ¼ ¼ þ1 γ1  1 p1 a1 T1 p1 ρ2 p1 γ1  1 p1

(11.105)

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11.12 The Shock Tube

331

In a similar fashion for the incident shock, Eqn. (11.92a) can be rewritten for the reflected shock: 2 p5 2γ1 Msr  ðγ1  1Þ ¼ γ1 þ 1 p2

(11.106)

Inserting Eqns. (11.92a), (11.105), and (11.106) into Eqn. (11.104), one finds γ1 þ1

p1

p5 γ1 1 þ 2  p2
 ¼ p2 1 þ γ1 þ1 p1 γ 1 p2

(11.107)

1

By using Eqn. (11.98) for p2 =p1 as a function of M1s and using the relationship T5 =T2 ¼ ðp5 =p2 Þðρ2 =ρ5 Þ, respectively, leads to    2 2 2γ1 M1s  ðγ1  1Þ ð3γ1  1ÞM1s  2ðγ1  1Þ p5 ¼ (11.108) 2 þ2 γ1 þ 1 p1 ðγ1  1ÞM1s 2 2 T5 ½2γ1 M1s þ ð3  γ1 Þ½ð3γ1  1ÞM1s  2ðγ1  1Þ ¼ 2 2 T1 ðγ1 þ 1Þ M1s

(11.109)

Equations (11.108) and (11.109) allow determining p5 and T5 , respectively, where M1s has to be first calculated using the implicit relationship, Eqn. (11.93). Thus, by choosing the initial pressure ratio in the driver and the driven chambers and initial pressure p1 and the initial temperature T1 , the conditions (p5 and T5 ) under which the shock-induced chemical reaction takes place is determined. A simple shock tube as shown in Fig. 11.16 is used so that the chemical reaction in region 5 is to be observed. For the desired temperature ratio T5 =T1 ¼ 7:94, what is the pressure ratio of the driver to driven gas is needed? Assume that γ ¼ 1:4 for all regions of gases. EXAMPLE 11.12

Solution – 2 2 þ ð3  γ1 Þ½ð3γ1  1ÞM1s  2ðγ1  1Þ T5 ½2γ1 M1s ¼ ¼ 7:94 2 2 T1 ðγ1 þ 1Þ M1s

One finds M1s ¼ 2:33 So that    2 2 2γ1 M1s  ðγ1  1Þ ð3γ1  1ÞM1s  2ðγ1  1Þ p5 ¼ ¼ 89:12 2 þ2 γ1 þ 1 p1 ðγ1  1ÞM1s

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332

11 Unsteady One-Dimensional Flows and Nonlinear Waves

Comments – If observation is to be made in region 2 for the same temperature ratio T5 =T1 , an exceedingly high p5 =p1 (on the order of hundreds of thousands) is needed, as □ shown in Example 11.11.

Problems Problem 11.1 Show all steps leading to Eqn. (11.15). Problem 11.2 Derive Eqns. (11.97) and (11.102). Problem 11.3 Equation (11.26) can be further simplified by introducing a thermodynamic function F ¼ F ð p; sÞ such that F≡

ðp

dp p0 ρa s

where p0 is a reference pressure and the subscript s indicates that entropy is being held constant. The s ¼ const: requirement is consistent with the earlier assumption r ⋅ s ¼ 0. Show that 

 ∂ ∂ þ ð u  aÞ ðu  F Þ ¼ 0 ∂t ∂x

By comparing Eqn. (11.36), F¼

2a γ1

Problem 11.4 Consider a piston in a long tube where the gas is initially at rest. If the piston is withdrawn to the left following the path line Xp ðtÞ, find (1) the slopes of the (right-running) characteristics originating from the piston surface, (2) the sonic velocity that the piston can attain, and (3) the piston velocity that would cause a vacuum. Compare these results with those in Problem 4.15 for steady flows. Problem 11.5 Repeat Example 11.5 with the advancing motion is twice as weak as the withdrawing motion, that is, uB ¼ up =4; uc ¼ up =2; uE ¼ up ; and up ¼ up . Problem 11.6 For the piston motion described in Fig. 11.8b, show (1) that the particle position at the passing of the ith (i.e., Ciþ ) characteristic is e xi ¼ 1 þ

iðγ þ 1Þ e up; 2n

(2) that the time the particle spends in the fan is

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Problems

333

 ð1þγÞ=ð1γÞ γ1 e Δet ¼ et n  et o ¼ 1 þ  1; up 2 (3) the total particle displacement through the expansion fan is xo ¼ Δe x¼e xn  e

ð γ þ 1Þ ð γ þ 1Þ e e u p or Δx ¼ up 2 2

Problem 11.7 Discuss the usefulness of Δp ¼ ρaΔu for (1) simple acoustic waves, (2) simple nonlinear continuous waves, and (3) shock waves. Problem 11.8 Consider the shock to be a left-running (C ) shock in Example 11.8. Show that the results are the same. Problem 11.9 In Example 11.5, compare the pressures at locations given by the following two sets of points: (1) 2, 5, and 8, and (2) 4, 5, and 6, and observe pattern of changes in their sound speeds. Problem 11.10 The wall boundary in Fig. 11.15 is moving with a constant speed ub ¼ a1 =4 to the right while the gas behind the incident wave is traveling at a speed equal to a1 =2. Find the pressure in each of regions 1, 2, and 3, given that the pressure in region 1 is p1 . Compare these results with those of Example 11.9. Problem 11.11 Compare the results of the incident wave in Example 11.9 for ub ¼ 0using exact solutions for traveling shock. Problem 11.12 For the results of Example 11.10, show that the slopes of characteristics at points 4, 7, 9, and 10 are in the following order: ðu10  a10 Þ > ðu9  a9 Þ > ðu7  a7 Þ > ðu14  a4 Þ. Explain this result. Problem 11.13 Expand the results of Example 11.10 to find the speed of the piston so that the reflected wave does not reach the piston. Compare the result with jup jescape ¼

2ao γ1

and explain the finding. Problem 11.14 The shock tube test as described in Fig. 11.16 has the following properties: T1 ¼ 300 K and T4 ¼ 500 K, γ4 ¼ 1:1, γ1 ¼ 1:4. To achieve T2 =T1 ¼ 5, find the needed p4 =p1 Problem 11.15 Derive Eqn. (11.18). Problem 11.16 For a general expansion wave, consider a piston motion in a cylinder containing initially quiescent gas. The piston trajectory is described by 1 xp ¼ αtp2 2

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334

11 Unsteady One-Dimensional Flows and Nonlinear Waves

where tp is the time counted from the instant when the piston is set in motion. The t-x plot of the piston movement is as shown. (a) Show that the particle location in the initially quiescent gas can be described by 

x  xp





  γþ1 α tp  a1 t  tp ; 2

   ¼ u  up t  t p ¼

where up is the speed of the piston and the subscript 1 designates the initially quiescent gas. x ≡ αx=a21 , (b) By defining the dimensionless parameters: et p ≡ αtp =a1 , et ≡ αt=a1 , and e show that 1 2 e x p ¼ et p 2 and 8 #12 9

"

= 1< γþ1 γþ1 2 2 et p ¼ et þ 1  ðγ  1Þet þ et  2γe 1þ x ; γ: 2 2 (c) Show that the speed of sound and the pressure in the gas, respectively, are a γ1 et p ¼1 a1 2   2γ p γ  1 γ1 et p ¼ 1 p1 2 (d) Show that the piston reaches the sonic and escape speeds, respectively, at et p ¼

2 γþ1

et p; escape ¼

2 γ1

And that the escape speed is reached at the location e xp ¼

2 ð γ  1Þ 2

Problem 11.17 Consider Xp ¼ A sin ωt in a long tube containing quiescent gas at t ¼ 0. Assume that A = a1/2ω, a1 being the speed of sound in the undisturbed gas. Downloaded from https:/www.cambridge.org/core. Columbia University Libraries, on 24 Jun 2017 at 04:20:25, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/9781316014288.012

Problems

335

Find the time and the location of the first shock formation, if the shock formation takes place at all. Explain your results by examining the slopes of the characteristics and compare the finding with those of Example 11.7. Problem 11.18 In a long tube, a moving normal shock wave is propagating against an M ¼ 2 air flow (at 1 atm and 400 K), bringing it to a full stop relative to a laboratory observer. Determine (1) the wave speed as observed by the laboratory observer, (2) the pressure ratio across the shock wave, and (3) the temperature in the stagnant air behind the wave and, by comparing this with the stagnation temperature in the incoming flow, explain what is found. Problem 11.19 A one-dimensional shock wave is moving normal to itself into stagnant air (at 300 K) causing a pressure ratio p2 =p1 ¼ 6, with 1 and 2 denoting the upstream and downstream regions of the shock, respectively. Find the Mach number behind the shock, Tt2 , pt2 and compare these results with those of a stationary normal shock with the same p2 =p1 . Problem 11.20 The piston in a long piston-cylinder device is suddenly withdrawn with a constant speed up (≪ a1 ), as shown in the following figure, where some representative incident characteristics (which form an expansion fan) are also shown. Region 1 is the undisturbed, uniform region, while region 2 is the uniform region downstream of the expansion fan. How would the first reflected characteristic, denoted by AB, curve in the non-simple region? u1 = 0

up

a1

Problem 11.21 The piston in a long piston-cylinder device is suddenly advanced with a constant speed up (≪ a1 ) to generate compression waves leading to a normal shock, as shown in the following figure. For the condition given in the figure, find the values of (a) up and (b) Vrs , the speed of the reflected shock.

up

p2 = 2p1 shock u1 = 0 u2 = up p1 = 1 atm T1 = 293 K 2 1 C+

t

3 C– 2

1 x

Problem 11.22 A long cylinder has two pistons that are suddenly set into relative motion at t ¼ 0þ with constant speeds, with uL and uR being the speed of the left and

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336

11 Unsteady One-Dimensional Flows and Nonlinear Waves

right pistons, respectively. At t ¼ 0, the gas between the two pistons is at rest (designated as the state 1) and at t ¼ ti the waves generated by the piston motion begin to interact and various flow regimes are as shown in the figure, where the dash lines are either compression or expansion waves depending on whether the piston motion is advancing or withdrawing. Find u4 , a4 , and p4 =p1 for the following scenarios, where up ≪ a1 – (1) (2) (3) (4)

uL ¼ uR ¼ up > 0 uR ¼ up > 0 and uL ¼ up < 0 uR ¼ 2up > 0 and uL ¼ up > 0 uR ¼ 2up > 0 and uL ¼ 0

up

uR

t 4 t = ti

3

2 1 x

Problem 11.23 A piston is set into an advancing motion in a very long tube where the initial state (denoted as state 1) of the gas is quiescent. The piston speed up ðtÞis such that a normal shock wave (speed = W1s ) forms ahead of it. For up ðtÞ ≪ a1

 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi h

i ffi (1) Show that Wa11s ¼

γþ1 4

(2) Show that W1s ¼ 12 form the shockÞ

up a1

þ
X

1þ

γþ1 4

up a1

2

speeds of the two characteristics that intersect to

(3) Find W1s for the piston motion described in Example 11.7. (4) The pressure rise when the shock is formed. Problem 11.24 A diaphragm in a long tube enables separation of two regions of unequal pressures. After the diaphragm is ruptured, a shock and an expansion waves form and propagate in opposite direction, as shown in the figure, where the flow regimes are also denoted by 1, 2, 3, and 4. The original contact surface is also set in motion. Find the difference between the shock wave and the contact surface speeds, (W1s  Wcs ) for the following scenarios – (1) Very weak shock (2) Very strong shock

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Problems

337

(3) Use the method of characteristics to show that u2 > 0. (4) Determine u2 =a1 and find its strong shock value. t=0 p4 > p1

t>0

4

Diaphragm P1

3

2 1 WCS WIS

Problem 11.25 A piston in a long tube is suddenly withdrawn at a constant speed, up , as shown in the figure. (a) Find the pressure on the piston face using the method of characteristics. (b) For up ≪ a1 , using binomial expansion to show that this pressure is same as that found using the momentum equation.

up

a0 , p0 , T0 u0 = 0

Problem 11.26 Consider the same device as in Problem 11.25, except that the piston is advancing with a speed up ¼ at, where a is the instant acoustic speed. Find the pressure at the piston face and determine when it becomes infinity. Problem 11.27 Show steps leading to Eqns. (11.44a) and (11.44b).

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12

Introduction to Inviscid Hypersonic Flows

Hypersonic flows are characterized by high Mach numbers and high temperatures resulting from strong shocks. They are characterized by the associated oblique shocks with small wave angles generated at the leading edge of the body, as typically occurs in the case of reentry vehicles. It is expected that due to the small shock wave angle, the interaction between the shock wave and the boundary can be significant. Due to the strong shocks, the temperature rises significantly to cause dissociation of gas molecules and reactions among them. The large increase in temperature due to shock or aerodynamic heating leads to challenges in the design of space vehicle. The gas mixture behind the shock wave cannot be assumed to be a perfect gas. Without treating the detailed aspects of the shock-boundary interaction and chemical reactions in such flows, this chapter focuses on the key effect of a large Mach number on the behaviors of shock and expansion waves in inviscid flows.

12.1 Relationships for Shock Waves in Hypersonic Flow A hypersonic flow is one in which Mach number is much greater than unity, and therefore the free-stream Mach number M1 is such that M1 ≡

V∞ ≫1 a∞

(12.1)

The shock relations obtained in Chapter 4 can be used by letting M1 ! ∞ and 2 ¼ M12 sin 2 θ ≫ 1. Thus, M1n sffiffiffiffiffiffiffiffiffiffiffi 2 2 M1n þ γ1 γ1 γ1 2 and M2n ! (12.2) M2n ¼ 2γ 2 ! 2γ 2γ γ1 M1n  1 p2 ¼ p1



2 2γM1n γ1 2γ M2 sin 2 θ  ! γþ1 1 γþ1 γþ1

(12.3)

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12.1 Relationships for Shock Waves in Hypersonic Flow

339

M1 >> 1 α=δ

θ

(M1

∝:θ=

γ +1 δ) 2

Increasing M1 (decreasing shock wave angle θ)

Figure 12.1 A flat plate at an angle of attack to free-stream hypersonic flow ðM1 ≫ 1Þ; variation of the shock wave angle, θ, with M1 and the angle of attack, α (¼ δ), shown schematically and mathematically.

T2 ¼ T1



2γ 2 2 1 þ γ1 2 M1n γ1 M1n  1 2γðγ  1Þ 2 2γðγ  1Þ 2 h i ! M1n ¼ M1 sin 2 θ 2 2 ðγþ1Þ2 2 ð γ þ 1 Þ ð γ þ 1 Þ 2ðγ1Þ M1n

(12.4)

pffiffiffiffiffiffiffiffiffiffiffi   2 ρ2 V1n M1n γRT1 ðγ þ 1ÞM1n γþ1 pffiffiffiffiffiffiffiffiffiffiffi ¼ ¼ ¼ ! 2 þ2 γ1 ρ1 V2n M2n γRT2 ðγ  1ÞM1n

(12.5)

   p 2  p1 2γ  2 2γ β≡ M1n  1 ! M2 sin 2 θ ¼ γþ1 γþ1 1 p1

(12.6)

To reduce the shock strength, hypersonic vehicles are designed to produce a small shock wave angle θ, which in turn requires a small deflection angle δ (i.e., slim body shapes). For small deflection angles (δ ≪ 1) and M12 sin 2 θ ≫ 1 Eqn. (4.33c) continues to hold. With sin θ ≈ θ, cos2θ ≈ 1, and tan δ ≈ sin δ ≈ δ, Eqn. (4.33c) simplifies to θ¼

γþ1 δ 2

(12.7)

This result indicates a linear relationship between the shock wave angle and the deflection (or wedge) angles. Inspection of Fig. 4.12 (for γ ¼ 1:4 and θ ¼ 1:2δ) provides support for the linearity expressed in Eqn. (12.7) as M1 ! ∞ over a range of δ, up to approximately δ ¼ 358 (i.e., δ ≈ 0:611), which is surprisingly larger than permitted by the assumption of δ ≪ 1. The result of Eqn. (12.7) is schematically shown in Fig. 12.1, where a flat plate is at an angle of attack to the free-stream flow at M1 ≫ 1. EXAMPLE 12.1

Access the validity of Eqn. (12.7) for δ ¼ 58 and 358 for an infinite free-stream Mach number M1 ! ∞, and for δ ¼ 308 and M1 ¼ 10. Solution – For δ ¼ 58 ≈ 0:0873, θ ≈ 68 ¼ 1:2δ. In fact for δ ¼ 358 ≈ 0:611, θ ≈ 42:088 ≈ 1:202δ, quite a pleasantly surprising result considering the 358 deflection is not quite a small angle. For other finite, but still large values of M1 , the linear relation

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12 Introduction to Inviscid Hypersonic Flows

between θ and δ is limited o a smaller range of δ, as can be seen from Fig. 4.12. For finite values of M1 , the deviations from the prediction using Eqn. (12.7) increase with decreasing M1 . Such a result is expected, because of the underlying assumption that M1 be infinity, and can be clearly visualized as the intercept on the θ axis increases with decreasing value of M1 . For example for M1 ¼ 10 and δ ¼ 308 , Fig. 4.12 gives, θ ≈ 39:258 ≈ 1:308δ, with Eqn. (12.7) under-predicting by □ approximately 9%. For the pressure coefficient across an oblique shock, Eqn. (4.57) results in Cp ≡

2 p 2  p1 2 4 sin 2 θ ¼ β¼ 2 2 γþ1 γM1 p1 γM1

(12.8)

For small deflection angles (δ ≪ 1 and sin θ ≈ θ ¼ ½γ þ 1δ=2 ≪ 1), Cp ≈ ðγ þ 1Þδ2 Recalling for small δ and moderate values of M1 , Eqn. (4.35) 0 1 2  B γM1 C p2  p1 2γ  2 M1 sin 2 θ  1 ¼ @qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ β¼ A⋅δ γþ1 p1 M12  1

(12.9)

(4.35)

can be used to obtain 2δ Cp ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M12  1

(4.59c)

where the positive and the negative signs denote, respectively, deflections toward the free stream (i.e., compression) and away from the free-stream flow (i.e., expansion). Comparing Eqns. (4.59c) and (12.9) reveals that for small values of δ: As M1 ! ∞, Cp is independent of M1 and Cp ∝δ for small δ and, as opposed to that for moderate and small supersonic values of M1 , Cp is a function of both δ and M1 .

12.2 Practical Considerations of Hypersonic Flow of Non-Perfect Gases Consider a hypersonic flow around a slender body of revolution, with a blunt nose/ leading edge, shown in Fig. 12.2. A slender body is of practical importance, as a thick body incurs tremendous pressure losses and is not at all appealing for use. In the nose region, the shock wave is detached and has the form of a normal shock. Away from the nose region, a bow shock forms, because the shock strength weakens in the direction away from the nose region (i.e., β≠ constant) along the bow shock, while further away the shock wave takes up the form of an oblique shock. Therefore, an entropy layer forms along the surfaces (see the discussion of Fig. 4.13). Since M1 ≫ 1, Eqn. (12.2) indicates that M2 ¼ 0:378, which causes the temperature in the nose

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12.2 Practical Considerations of Hypersonic Flow of Non-Perfect Gases

341

gth

en

De M1 >> 1

h

gs

sin

a cre

tr ks oc

oblique shock in the far field

Figure 12.2 A hypersonic flow around a slender body of revolution, with a blunt nose/leading edge, with the shock transitioning from a bow shock in the nose region to an oblique shock in the far field.

region to increase to nearly stagnation value. For M1 ¼ 20, T2 ¼ 77:78 T1 (from Eqn. (12.4)) which at an altitude of 30 km based on the ideal gas properties and γ ¼ 1:4, T2 is approximately 17,000 K, compared to the stagnation value of 17,600 K on the surface of the nose. Such high temperatures have several effects, including dissociation and ionization of gases (air in the atmosphere) and chemical reactions. For example, the dissociation of O2 and N2 in air leads to formation of NO (nitric oxide), which is a non-equilibrium process. Ablation of the surface materials may also occur under such harsh environments. High temperatures lead to strong thermal radiation from both gases and surfaces. Because ionization and formation of new species require energy, and radiation emits heat, the temperature of the newly established equilibrium mixture further downstream of the shock wave would be lower, although still high. As a result of ionization and dissociation, the assumption for ideal gases is no longer valid. Furthermore, determining the gas composition in this region (and in the hypersonic boundary layer where the low gas velocity causes a high “recovery” temperature near the stagnation value) requires non-equilibrium thermodynamic and chemical models. Validation of these models is as challenging as obtaining experimental data for gaseous species in such environments (traditional sampling probes or non-intrusive optical methods) and is not a trivial matter. Consequently, thermal and physical properties (such as cp , c–v , and γ) of the gases are not easily determined because of the change in gas composition across the shock wave, which also varies along the shock wave with non-uniform strength along its surface. Knowledge of how internal energy storage depends on temperature can help shed light on the value of cv . The dissociation results from the increasingly vigorous vibration between atoms that breaks down their bonds. The high temperature therefore has significant effects on the heat capacity cv in the following manner. Combining the result for cv at moderate temperatures from Chapter 1 (i.e., c–v ¼ ξR=2, where ξ is the number of degrees of freedom of the molecular translation and rotation) and the contribution from vibration leads to c–v ¼

ξ R þ cv;vib 2

(12.11)

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12 Introduction to Inviscid Hypersonic Flows

1.0 Cv,vib R

Cv,vib /R

0.8

=

2

Θvib/2T sinh(Θvib/2T)

0.6

Figure 12.3 Variation of cv;vib , the constant-volume heat capacity due to vibration of the molecule, with temperature.

0.4

0.2

0.0

0

1

2 T/Θvib

3

4

where cv;vib represent the constant-volume heat capacity due to vibration of the molecule. It is useful to note the “vibrational temperature,” Θvib , which is the characteristic temperature above which the contribution from the vibrational motion to the total heat capacity c–v is not negligible. At high temperatures the electrons are also “excited”; however, at very high temperatures, the energy level of electrons varies only very slightly and does not contribute significantly to c–v . The values of Θvib for O2, N2, and NO are 2,270 K, 3,390 K, and 2,740 K, respectively (Vincent and Kruger, 1975). The temperature dependence of cv;vib on temperature is c–v;vib

 ¼R

Θvib =T sinhðΘvib =2T Þ

2 (12.12)

Results of Eqn. (12.12) are shown in Fig. 12.3. It can be seen that as the temperature approaches 17,000 K, the mixture of the diatomic gases of interest (O2, N2, and NO; ξ ¼ 5) possesses a constant-volume specific heat equal to ξ ξþ2 R c–v ¼ R þ R ¼ 2 2

(12.13)

Assuming that the gas behind the shock wave comprises of mainly O2, N2, and NO, the specific heat ratio is thus γ¼

R cp ξþ2 ¼1 ¼ 2 c–v ξþ2 2 R

(12.14)

At moderate temperatures, values of c–v vary and increase with temperature, resulting in a temperature regime of variable γ. In the flow around the body shown in

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12.2 Practical Considerations of Hypersonic Flow of Non-Perfect Gases

343

Fig. 12.2, the value of γ also varies from one region to another because of the nonuniform shock strength and temperature jump across the shock wave system (for the same reason for the entropy layer). Therefore it is not unusual to choose an effective value of γ (γeff ) for the convenience of solving the problem, bypassing the difficulty of ascertaining the detailed and spatially varying composition and temperature of the gas undergoing dissociation reactions. For diatomic gases (of which air is one), the effective value lies between unity and 1.4. In the limit of M1 ¼ ∞, γ ¼ γeff ¼ 1. Under these conditions, Eqn. (12.7) becomes θ¼δ

(12.15)

This result has several implications. First, the shock wave is inclined at the same angle as the two-dimensional wedge. In reality, θ is slightly larger than δ, making the interaction between the boundary layer and the shock wave significant. For the qualitative analysis, further consider the results for M1 ! ∞ shown in Fig. 4.11 (where a constant γ ¼ 1:4 is used and in the limit of M1 ∞, θ ≈ 1:2δ). Figure 4.11 indicates that θθ decreases with increasing M1 and that for the constant value γ the shock wave moves toward the surface as M1 is increased. Equation (12.15) suggests that in hypersonic flows, such a movement with increasing M1 brings the shock wave to the surface; such a movement is illustrated in Fig. 12.1, over a thin flat plate. The shock wave-boundary layer interaction is outside the scope of this chapter; for this topic one is referred to, for example, Anderson (1989) and Park (1990). With M1 ¼ ∞ and γ ¼ γeff ¼ 1, Eqns. (12.9a) and (12.9b) for a flat plate at an angle of attack α becomes Cp ¼ 2 sin 2 θ ¼ 2 sin 2 α

(11.16a)

Equation (11.16a) is the so-called sine-squared law for hypersonic flow. Because α ¼ δ ≪ 1 and sin θ ≈ θ ¼ ½γeff þ 1δ=2 ≪ 1, it becomes Cp ≈ ðγ þ 1Þδ2 ¼ 2θ2 ¼ 2α2

(11.16b)

It is of interest to be able to obtain values of Cp on the surface of a blunt body in hypersonic flow (Fig. 12.4). For the portion of the surface of the blunt body that is normal to the streamline, a normal shock wave forms. The flow between the normal shock wave and the body decelerate to zero velocity, forming a stagnation point at the surface, where the pressure is the stagnation pressure (p2 ¼ pt2 ) behind the shock and the surface pressure coefficient is the maximum on the body, denoted as Cpmax . Let the subscript 2 designate the state behind the shock. Thus Cp;max ¼

ðpt2  p1 Þ 1 2 2 ρ1 V1

(12.17)

The pressure coefficient elsewhere can be expressed as (Thompson, 1972) Cp ¼ Cp;max sin 2 θ

(12.18a)

or

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344

M1

12 Introduction to Inviscid Hypersonic Flows

∝ α

Figure 12.4 Schematic showing the local angle of attack, θ, formed by the tangent to the airfoil surface and the flow direction, and the angle of attack formed by the chord line, α, in hypersonic flow.

θ

P

Cp ¼ sin 2 θ Cp;max

(12.18b)

where θ is now taken as the local angle of attack, or the inclination angle to the freestream flow, of the surface location away from the stagnation. Such a local angle of θ (e.g., of point P) is illustrated in Fig. 12.4. EXAMPLE 12.2

Estimate the Mach number and static temperature of the post-shock region for the flat-plate configuration shown in Fig. 8.3 with α ¼ 358 . The given flow conditions are M1 ¼ 20 and T1 ¼ 220 K. Solution – For the free stream, the isentropic relationship Eqn. (4.6c) can be used and for γ ¼ 1:4 T1t γ1 2 M1 ¼ 1 þ 0:2  202 ¼ 81; T1t ¼ 81  220K ¼ 17; 820 K ¼ 1þ 2 T1 Oblique shock theory Immediately downstream of the shock wave where dissociation/ionization/reaction have not occurred yet, γ ¼ 1:4. From Fig. 4.12 θ ¼ 458 , then M1n ¼ M1 sin θ ¼ 14:14. The shock relationship Eqn. (4.22d) provides

 γ1 2γ 2 2 2 2 1 þ M sin θ M sin θ  1 1 1 2 γ1 T2 h i ¼ ¼ 39:82 ðγþ1Þ2 T1 2 2 M sin θ 2ðγ1Þ

1

Or using M12 sin 2 θ ≫ 1 T2 2γðγ  1Þ 2 ¼ M1 sin 2 θ ¼ 38:89 T1 ðγ þ 1Þ2 Therefore T2 ≈ 8; 556  8; 760 K. The shock relationship, Eqn. (4.22a), gives M2n as 2 M2n ¼

2 2 M1n þ γ1 2γ 2 γ1 M1n

1

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12.2 Practical Considerations of Hypersonic Flow of Non-Perfect Gases

345

M2t ¼ 0:3828 rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi rffiffiffiffiffiffi T1 T1 1 8 ¼ 2:241 M2t ¼ M1t ¼ M1 cos θ ¼ 20  cos 45  39:82 T2 T2 M2 ¼ 2:2735 It should be noted that the estimation of T2 is based on γ ¼ 1:4, which is itself contradictory as 8; 556 K exceeds the vibrational temperature for typical species present behind the shock wave Θvib ¼ 2,270 K, 3,390 K, and 2,740 K for O2, N2, and NO, respectively. Hypersonic theory For M1 ¼ 20, assume that M1 ! ∞, then γeff ! 1 in this region. Further downstream of the shock wave, a new equilibrium exists and γeff ! 1. As a first approximation, one may still assume an ideal gas in this region and that the gas velocity does not change due to a new equilibrium. Also θ ¼ δ ¼ α ¼ 358 . One readily find M1n ¼ M1 sin θ ¼ 20  sin 358 ¼ 11:47. However, both T2t γ 1 2 M2 ¼ 1 ¼ 1 þ eff 2 T2

M2n

and

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi γeff  1 ! !0 2γeff

are apparently not the case because M2n ¼ M2 sin θ would imply that θ ¼ 0 and a stagnation flow downstream of the shock, violating the given condition θ ¼ 358 . One can use the l’Hospital rule for the following result " 2 # 2 M1n þ γ1 1 1 1 2 ¼ 0:08718 M2n ¼ lim 2γ 2 ¼ 2 or M2n ¼ ¼ γ!1 M1n 11:47 M1n γ1 M1n  1 M2t ¼ M1t

rffiffiffiffiffiffi T1 T2

Also θ ¼ δ ¼ α ¼ 358 . Assume that the momentum flux in the tangential direction along the shock wave remains the same. Then following Eqn. (4.21d), V2t ¼ V1t

and

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346

12 Introduction to Inviscid Hypersonic Flows


T2 ¼ T1


 2γ 2 2 2 2 1 þ γ1 M sin θ M sin θ  1 1 1 2 γ1 h i ¼ 26:5227 ðγþ1Þ2 2 2θ sin M 1 2ðγ1Þ

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi rffiffiffiffiffiffi T1 T1 1 8 M2t ¼ M1t ¼ M1 cos θ ¼ 20  cos 35  ¼ 3:1812 26:5227 T2 T2 M2 ¼ 3:1824 One can see that the hypersonic theory (γeff ! 1) predicts a smaller T2 and □ a larger M2 than the oblique shock theory assuming γ ¼ 1:4. Fine the pressure coefficients for a flat plate in a hypersonic air flow at angles of attack of 58 and 358 , respectively.

EXAMPLE 12.3

Solutions – For α ¼ 58 ¼ 0:08731, Cp ≈ 2α2 ¼ 0:0152 (also Cp ¼ 2 sin 2 α ¼ 0:0152). For α ¼ 358 ¼ 0:611, which is not much smaller than unity, Cp ¼ 2 sin 2 α ¼ □ 0:658 (2α2 would give a value of 0.746). Find an expression for Cpmax and Cp of a blunt body in hypersonic flow. The angle θ is as shown in the figure.

EXAMPLE 12.4

M1

θ

∝

θ

Solution – The stagnation pressure behind the normal shock wave in Eqn. (11.17), pt2 , is related to the free-stream stagnation, pt1 , by Eqn. (4.19d): pt2 ¼ pt1

" 1

#γ=ðγ1Þ " #1=ðγ1Þ γþ1 2 1 2 M1 2γ γ1 2 2 þ γ1 2 M1 γþ1 M1  γþ1

" ≈ ¼

1

#γ=ðγ1Þ " #1=ðγ1Þ γþ1 2 1 2 M1 2γ 2 2 þ γ1 γþ1 M1 2 M1

ð Þ

By using the ideal gas law, p ¼ ρRT, Cp;max can be rewritten as

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12.3 Lift and Drag in Hypersonic Flow

ðpt2  p1 Þ ðpt2  p1 Þ ¼ ¼ 1 2 V12 γp1 2 ρ1 V1

Cp;max ¼

2



γRT1

pt2 p1

347





2 2 pt2 2 ≈  γM1 p1 γM12 γM12



pt2 pt1 2 ¼ pt1 p1 γM12

Substituting (*) and the isentropic relation for pt1 =p1 , with M1 ≫ 1, into this expression yields " Cp;max ¼

1

#γ=ðγ1Þ  γþ1 2 2 M1 2 γ þ γ1 2 M1



γþ1 ¼ γ

2γ M2 þ1 1

1=ðγ1Þ     γ  1 2 γ=ðγ1Þ 2 M1 2 γM12

γ  γ1 1 γ þ 1 γ1 4

which is independent of M1 for M1 ≫ 1. Cp ¼ Cp;max sin 2 θ Forγ ¼ 1:4, Cp ¼ 1:839 sin 2 θ Forγ ¼ γeff ¼ 1, Cp ¼ 2 sin 2 θ which recovers the result shown by Eqn. (12.16a) for a flat plate; for the correct □ result, θ is the local angle relative to the free stream.

12.3 Lift and Drag in Hypersonic Flow An engineering method of calculating pressure on surfaces in hypersonic flows is the Newtonian impact theory. The theory requires that fluid particles in the free stream only impact the frontal surface area of the object and that they cannot flow around the object. The latter requirement appears to be reasonable as high-speed flows have large inertia that prevents turning. For illustration, consider an infinitely thin flat plate at an angle of attack α to a hypersonic stream, as shown in Fig. 12.5. It is noted that θ ¼ α everywhere on the lower surface of a flat plate. The control volume in Surface area = A A Sin α

V1 α

Figure 12.5 Schematic of an infinitely thin flat plate at an angle of attack α to a hypersonic stream, showing relevant parameters for the Newtonian impact theory.

Control surface

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348

12 Introduction to Inviscid Hypersonic Flows

Fig. 12.5 is drawn such that there is no flow over the upper surface, as there is zero inlet area over the upper surface. As a consequence, pU ¼ p1 and CpU ¼ 0. In contrast, the flow impacting the lower surface enters the control surface through an area equal to A sin α. Applying the Reynolds transport theorem, Eqn. (2.8), for the inviscid flow under steady state condition leads to

X

ð

ð

~¼ ðpL  p1 Þ ⋅ dA

~ F ¼ CS


 ~ ~ ρV ~ ⋅ dA V

CS

For the pressure force acting normal to the flat plate, pL  p1 ¼ ρ1 V12 sin 2 α

(12.19)

which in turn leads to the sine-squared law for the lower surface: CpL ¼ 2 sin 2 α

(12.20)

In arriving at Eqn. (11.20), conditions of M1 ! ∞, the presence of a shock wave and γ ¼ γeff (i.e., high-temperature thermodynamics) are not invoked. Furthermore, Eqn. (12.20) suggests the independence of pressure coefficient on M1 . The modified Newtonian impact theory can be obtained by considering shock relations, where γ affects the pressure ratio across a shock wave. Thus, Eqn. (12.8) essentially represents the modified theory – for a flat plate, θ ¼ α and Cp ¼ ½4=ðγ þ 1Þ sin 2 α < 2 sin 2 α for γ > γeff as already demonstrated in Example 11.4. For a body of an arbitrary shape, θ varies from one surface location to another and thus θ≠α. By using the local angle of attack θ in place of α in Eqn. (12.20), CpL is independent of both γ and M1 for M1 ! ∞. The independence of γ is advantageous because the composition arising from dissociation of gases need not be known. Equation (12.20) can thus be extended to different gas species in the free stream. Together with the assumptions and results of Eqns. (12.16a) and (12.16b), one expects the accuracy of Newtonian impact theory to improve as M1 and θ (α in the case of a flat plate) increase. Such improvement of the Newtonian impact theory might be expected since, as M1 is increased toward ∞, the very large flow inertia renders going around the object increasingly difficult, consistent with the assumption of the theory. Increasing both M1 and θ causes the backside of the object not to be reached by the flow, thus becoming a shadow region. According to shock theory, increasing both M1 and θ increases shock strength and the temperature behind the shock, leading to the larger degrees of dissociation and pushing the value of γ toward 1. The coefficient 2 in Eqn. (11.20) is in strikingly good agreement with that of Eqn. (11.16a). This agreement for M1 ! ∞ is worth noting as the Newtonian impact theory does not consider the high-temperature effects on gas properties shown by Eqns. (12.11) through (12.14), as the Reynolds transport theorem is concerned with the rate of momentum change that does not depend on composition of the gas.

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12.4 Hypersonic Similarity

349

Following similar procedures for the thin airfoils in Section 4.10, the lift coefficient for the thin flat-plate airfoil (where the angle of attack is equal to the flow deflection angle, α ¼ δ) would generate lift and drag coefficients as   CL ¼ CpL  CpU cos α ¼ 2 sin 2 α cos α ð ≈ 2α2 for small αÞ (12.21)   CD ¼ CpL  CpU sin α ¼ 2 sin 3 α ð ≈ 2α3 for small αÞ

(12.22)

Comparison with lift and drag coefficients for thin airfoil at small angles of attack in supersonic flows, Eqns. (4.59a) and (4.59b) reveals (1) the lift and drag coefficients for thin supersonic airfoil depend on the free-stream Mach number and vary with α and α2 , respectively, while (2) the lift and drag coefficients in hypersonic flows are independent of the free-stream Mach number and are proportional to α2 and α3 , respectively. It is left as an end-of-chapter problem (Problem 12.1) to show that CL for a flat plate in hypersonic flow reaches a maximum of approximately 0.770 at α ≈ 558 and that the CL =CD ratio deteriorates rapidly as α is increased. It can also be shown that Newtonian theory predicts the drag coefficient for a large-aspect ratio circular cylinder, with the span placed perpendicular to the flow, to be 4/3. Similarly, for a sphere, the drag coefficient is 1. The process of arriving at these drag coefficients is left as an end-of-chapter problem (Problem 12.2).

12.4 Hypersonic Similarity Recalling the similarity variables for subsonic and supersonic flows described in Chapter 9, one might ask for the existence of similarity variables for hypersonic flow over slender bodies (characterized by, for example, small deflection angle δ). The following steps demonstrate how such a similarity variable is obtained. First, consider the oblique shock relation governing the deflection and shock wave angles applies for hypersonic flows: tan δ ¼ 2cotθ

M12 sin 2 θ  1 þ cos2θÞ þ 2

M12 ðγ

(4.27)

By taking advantage of smaller values of δ and θ, for which tan δ ≈ δ, sin θ ≈ θ, cos2 θ ≈ 1, and cos θ ≈ 1, Eqn. (4.27) can be rearranged to become   M12 θ2  1 2 (12.23) δ¼ θ M12 ðγ þ 1Þ þ 2 Rearranging Eqn. (11.23) leads to  2   2  M 1 ð γ þ 1Þ M 1 ð γ þ 1Þ 2 2 þ 1 δθ ≈ M1 θ  1 ¼ δθ 2 2

(12.24)

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12 Introduction to Inviscid Hypersonic Flows

where M12 ðγ þ 1Þ=2 ≫ 1 is valid but M12 θ2 ≫ 1 cannot be assumed because θ is small. Note that if constants 1 and 2 are neglected in Eqn. (12.23), Eqn. (12.7) is recovered: θ ¼ δðγ þ 1Þ=2. Equation (11.24) is then cast in the form of the quadratic equation: 2

θ γþ1 θ 1   2 2¼0 δ 2 δ M1 δ

(12.25)

which has the solution: θ γþ1 ¼ þ δ 4

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

γþ1 2 1 þ 2 2 4 M1 δ

(12.26)

Equation (12.26) indicates that the ratio of the shock wave angle to the deflection angle is only a function of the product M1 and δ; that is, θ=δ= f ðM1 δÞ. In the explicit form, θ ¼ f ðM1 ; δÞ, which is not similar to the weak solution for oblique shock waves described in Chapter 4 by Eqn. (4.34). One now can relate the pressure ratios shown in Eqns. (12.3) with θ given by Eqn. (12.26) to the M1 δ as the following p2 2γ 2γ 2γ γðγ þ 1Þ 2 2 M12 sin 2 θ ≈ M12 θ2 ¼ þ M1 δ ¼ γþ1 γþ1 4 p1 γ þ 1 s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

γþ1 2 1 2 2 þγM1 δ þ 2 2 4 M1 δ

(12.27a)

Similarly, T2 2γðγ  1Þ γðγ  1Þ 2 2 γðγ  1Þ 2 2 M1 δ þ M δ ¼ þ 4 ð γ þ 1Þ 1 T1 ð γ þ 1Þ 2

s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

γþ1 2 1 þ 2 2 4 M1 δ

Equation (12.5) remains unchanged:   ρ2 V1n ðγ þ 1ÞM12 sin 2 θ γþ1 ¼ ¼ ≈ γ1 ρ1 V2n ðγ  1ÞM12 sin 2 θ þ 2

(12.27b)

(12.27c)

while Eqn. (12.6), the shock strength, becomes p2  p1 γ  1 γðγ þ 1Þ 2 2 þ M1 δ þ γM12 δ2 β≡ ¼ γþ1 4 p1

s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

γþ1 2 1 þ 2 2 4 M1 δ

(12.27d)

It is of significance to note that in Eqns. (11.28a) through (11.28d), the common factor M1 δ determines the values of these ratios. Therefore K ≡ M1 δ

(12.28)

is the hypersonic similarity parameter. Equations (11.27a) through (11.27d) can now be rewritten as

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12.5 Hypersonic Expansion

p2 2γ γðγ þ 1Þ 2 þ K þ γK 2 ¼ 4 p1 γ þ 1

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

γþ1 2 1 þ 2 4 K

T2 2γðγ  1Þ γðγ  1Þ 2 γðγ  1Þ 2 K þ K ¼ þ 4 ð γ þ 1Þ T1 ð γ þ 1Þ 2 γ  1 γðγ þ 1Þ 2 þ K þ γK 2 β¼ γþ1 4

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

γþ1 2 1 þ 2 4 K

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

γþ1 2 1 þ 2 4 K

351

(12.29a)

(12.29b)

(12.29c)

The pressure coefficient, shown in Eqn. (11.9a), can be expressed as s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi " #

2 p2  p1 2 γ  1 γðγ þ 1Þ 2 2 γþ1 2 1 2 2 þ M1 δ þ γM1 δ ¼ þ 2 2 Cp ≡ 4 4 γM12 p1 γM12 γ þ 1 M1 δ which can be further reduced to 2 γþ1 þ Cp ¼ 2δ2 4 4

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3

γþ1 2 1 þ 25 4 K

(12.29d)

Equation (12.29d) shows that the pressure coefficient on a slender body in hypersonic flow is proportional to δ2 for a given value of K ≡ M1 δ, or Cp ¼ f ðK; γÞ δ2

(11.30)

The result of Eqn. (11.30) is in contrast to the pressure coefficient for moderately supersonic values of M1 , expressed by Eqn. (4.59c): 2δ Cp ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðþ for oblique shocks; for moderately supersonic values of M1 Þ M12  1 (4.59c)

12.5 Hypersonic Expansion Consider the centered Prandtl-Meyer expansion of a hypersonic flow (M1 ! ∞) around a corner through an angle of δ, as shown in Fig. 12.6. Adopting the same notations as for supersonic expansion, ðMÞ is the Prandtl-Meyer function: sffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi γþ1 γ1 tan 1 ðM2  1Þ  tan 1 M2  1 (4.47c)  ðM Þ ¼ γ1 γþ1 As M1 ! ∞, Eqn. (4.47c) becomes

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352

12 Introduction to Inviscid Hypersonic Flows

M1 >> 1

φ1−2

Expansion fan

P1 , T1

M > 2 M

δ

P2 ,

1

T2

Figure 12.6 Hypersonic flow expansion around a corner.

sffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffi γþ1 1 γ  1 tan M  tan 1 M  ðMÞ ¼ γ1 γþ1

(12.31)

The trigonometric identity for the tan 1 function provides tan 1 M ¼

π 1  tan 1 2 M

(12.32a)

Similarly, tan

1

sffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffi γ1 π 1 γ þ 1 1 M ¼  tan γþ1 2 γ 1M

(12.32b)

For ε ≪ 1, the series expansion 1 1 1 tan 1 ε ¼ ε  ε3 þ ε5  ε7 þ … (12.32c) 3 5 7 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi can be used for 1=M ≪ 1, while ðγ þ 1Þ=ðγ  1Þ is of order unity for expansion, in Eqns. (12.32a) and (12.32b). By manipulating Eqns. (11.33a) and (11.33b), substituting the results into Eqn. (12.31), and then retaining the first-order terms yields: "sffiffiffiffiffiffiffiffiffiffiffi # γþ1 π 2 1 1  (12.33)  ðM Þ ¼ γ1 2 γ  1M Recalling that the turning angle δ is related to the Prandtl-Meyer function by   2 1 1 δ ¼ Δ 12 ¼  ðM2 Þ   ðM1 Þ ¼  (12.34) γ  1 M1 M2 For Prandtl-Meyer expansion, M2 > M1 and δ > 0. The expansion fan angle, , according to Eqn. (4.49) is 12 ¼ Δ 12 þ sin 1

1 1 1 1  sin 1 ¼¼  ðM2 Þ   ðM1 Þ þ sin 1  sin 1 M1 M2 M1 M2

Because M2 > M1 and M1 ! ∞, both δ (for no flow separation) and 12 are expected to be small. For M1 ¼ 13 and M2 ¼ 15, δ ¼ 2:948 and 12 ¼ 3:538 . Assuming no flow separation due to the hypersonic turning around the angle δ, the isentropic pressure relation holds. For both M2 and M1 ! ∞, Eqn. (4.4) becomes

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Problems

2 1 þ γ1 2 M1

p2 ¼ p1

!γ=ðγ1Þ

2 1 þ γ1 2 M2

2γ=ðγ1Þ M1 ¼ M2

353

(12.35)

By rewriting Eqn. (12.34) as M1 γ1 M1 δ ¼1 2 M2

(12.36)

and substituting this result into Eqn. (12.35) yields p2 ¼ p1



2γ=ðγ1Þ γ1 M1 δ 1 2

(11.37)

The variable M1 δ is defined as the hypersonic similarity parameter for expansion: K ≡ M1 δ. Thus p2 ¼ p1



2γ=ðγ1Þ γ1 K 1 2

(12.38a)

Other ratios can be similarly found to be γ1 T2 1 þ 2 M12 ¼ ¼ T1 1 þ γ1 M22 2

β¼

p2  p1 ¼ p1



2γ γ1 K 1 2



2γ=ðγ1Þ γ1 K 1 1 2

" #

2γ=ðγ1Þ 2 p2  p1 2 γ1 K ¼ 1 1 Cp ¼ 2 γM12 p1 γM12 " #

2γ=ðγ1Þ 2δ2 γ1 ¼ 2 K 1 1 2 γK

(12.38b)

(12.38c)

(12.38d)

Similar to the pressure coefficient found for oblique shock, Eqn. (12.38d) can be cast in the functional form as Cp ¼ f ðK; γÞ δ2

(12.39)

Problems Problem 12.1 Use the Newtonian impact theory to find the angle of attack for which the lift coefficient is maximum for a thin flat plate in a hypersonic free stream. What are the values of CL =CD and CD when CL is maximum? To visualize these results, plot these parameters as a function of α.

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354

12 Introduction to Inviscid Hypersonic Flows

Problem 12.2 Use the Newtonian impact theory to find the drag coefficient for a long cylinder (with span placed perpendicular to the flow) and a sphere in hypersonic flow. Problem 12.3 Show that the pressure coefficient at the stagnation point of a blunt body in hypersonic air flow, Cp;max is equal to 2 by using the Newtonian impact theory and that Cp Cp;max

¼ sin 2 θ

Problem 12.4 By using the normal shock theory to find the pressure at the stagnation point of a blunt body in hypersonic air low, show that Cp;max ¼ 4=ðγ þ 1Þ

Cp ¼

4 sin 2 θ γþ1

When this result is used for Problem 11.3, a modified Newtonian impact theory is obtained, where the value of γ is a function of temperature behind the normal shock wave and Cp;max ¼ 1:667 for γ ¼ 1:4 (no dissociation of gases due to shock heating) and 2 for γ ¼ 1 (vibration modes of gas molecules fully excited); the realistic value of Cp;max should fall between 1.667 and 2. A modified theory assuming γ ¼ 1:4 predicts that Cp ¼ 1:839 sin 2 θ. Problem 12.5 Use the modified Newtonian theory shown in Problem 12.4 to obtain the drag force per diameter on a sphere flying at hypersonic speed. Also show that the drag coefficient is

θ

dθ

dr M1

P1 T1 V1

∝

r

r0

d0 p∝

1 CD ¼ Cp;max 2 Problem 12.6 Show that a hypersonic flow over a two-dimensional wedge with a small turning angle, δ, results in the downstream Mach number

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Problems

355

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 M2 ≈ γðγ  1Þ δ (Note that this result is not valid for γ ¼ γeff ! 1. The value of M2 is strongly dependent on γ, pointing to the importance of knowing the values of cp , cv , and γ for hypersonic approximations.) What can be said about the Mach number variation in a hypersonic flow over a blunt body, where a bow shock forms and the turning angle diminishes in the downstream direction away from the stagnation point region?

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APPENDIX A

Universal Physical Constants

Table A.1 Universal Physical Constants h ¼ Planck’s constant ¼ 6:26192  1034 J ⋅ s k ¼ Boltzmann’s constant ¼ 1:38054  1023 J=K NA ¼ Avogadro’s number ¼ 6:02252  1023 1=mol Rˆ ¼ universal gas constant ¼ 8:3143 kJ=kmol ⋅ K

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APPENDIX B

Properties of Some Ideal Gases

Table B.1 Thermodynamic Constant for Common Gases*

Air Argon Carbon Dioxide Carbon Monoxide Helium Hydrogen Methane Nitrogen Oxygen Water

Chemical Symbol

Molecular Weight

cp ðJ=kg ⋅ KÞ at 298.15 K

cv–ðJ=kg ⋅ KÞ at 298.15 K

γ

RðJ=kg ⋅ KÞ

Ar CO2 CO He H2 CH4 N2 O2 H2O

28.96 39.95 44.01 28.01 4.00 2.02 16.04 28.01 32.00 18.02

1004.0 520.3 841.8 1041.3 5192.6 14209.1 2253.7 1041.6 921.6 1872.3

716.5 312.2 652.9 744.5 3115.6 10084.9 1735.4 744.8 661.8 1410.8

1.400 1.667 1.289 1.398 1.667 1.409 1.299 1.400 1.393 1.327

287.06 208.13 188.92 296.83 2077.03 4124.18 518.35 296.80 259.83 461.5

* Sources – (1) Handbook of Chemistry and Physics, 74th Edition, CRC Press (1993); (2) Introduction to Thermodynamics, Classical and Statistical, 3rd Edition, R.E. Sonntag and G.J. Van Wylen, John Wiley & Sons (1991).

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APPENDIX C

Tables for Isentropic and Prandtl-Meyer Expansion Flows

Note – The numerical values of the table can be directly obtained from the following equations. They are tabulated for γ ¼ 1:4 convenience. (The subscript t denotes stagnation properties; the superscript * denotes critical condition for M ¼ 1.) T ¼ Tt p ¼ pt ρ ¼ ρt





γ1 2 M 1þ 2

γ1 2 M 1þ 2

1þ

γ1 2 M 2

1 (4.6c)

γ=ðγ1Þ (4.7)

1=ðγ1Þ (4.8)

sffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi γ þ 1 1 γ  1 tan  ðM Þ ¼ ðM2  1Þ  tan1 M2  1 ðPrandtl-Meyer FunctionÞ γ1 γþ1 (4.47f) A 1 ¼  A M



2 γþ1



γ1 2 M 1þ 2

ðγþ1Þ=2ðγ1Þ (5.14)

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360

Appendix C Tables for Isentropic Flows

γ = 1.4 Isentropic Temp Ratio Mach Number T/Tt

Isentropic Pressure Ratio p/pt

Isentropic Density Ratio ρ/ρt

Isentropic Area Ratio A/A*

Prandtl-Meyer Function v (degrees)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9

1 0.99972005 0.998880806 0.997484077 0.995532872 0.993031385 0.989984975 0.986400146 0.982284517 0.977646787 0.972496703 0.966845014 0.960703428 0.954084563 0.947001894 0.939469698 0.931503001 0.923117512 0.914329565 0.905156057 0.895614383 0.885722374 0.875498229 0.864960453 0.854127793 0.843019175 0.831653642 0.820050293 0.808228222 0.796206467 0.784003951 0.771639428 0.75913144 0.746498261 0.733757861 0.720927861 0.708025494 0.695067572 0.682070453 0.669050012 0.656021618 0.643000108 0.629999769 0.617034321 0.604116905 0.591260072

1 0.999800028 0.999200448 0.998202266 0.996807154 0.995017448 0.992836132 0.990266835 0.987313813 0.983981938 0.980276677 0.976204074 0.971770732 0.966983786 0.961850883 0.956380153 0.950580183 0.944459989 0.938028987 0.931296964 0.924274043 0.916970659 0.90939752 0.901565579 0.893486002 0.885170134 0.876629471 0.867875626 0.858920296 0.849775239 0.840452235 0.830963068 0.821319487 0.811533189 0.801615788 0.791578791 0.781433577 0.771191372 0.760863231 0.750460018 0.739992386 0.729470763 0.718905336 0.708306037 0.697682531 0.687044203

28.94213019 14.48148593 9.665910065 7.261609645 5.82182875 4.864317646 4.182399799 3.672738634 3.277926451 2.96352 2.7076021 2.495562452 2.31728731 2.165553576 2.035065262 1.921851275 1.822875752 1.735778282 1.658696166 1.59014 1.5289048 1.474005374 1.42462855 1.380097348 1.33984375 1.303387758 1.270321102 1.240294439 1.213007209 1.188199506 1.165645539 1.145148306 1.126535244 1.109654642 1.094372679 1.080570939 1.06814435 1.056999429 1.047052813 1.03823 1.030464279 1.023695814 1.017870853 1.012941048 1.008862865

− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −

1 0.999920006 0.999680102 0.999280518 0.998721636 0.998003992 0.997128271 0.996095306 0.994906081 0.99356172 0.992063492 0.990412804 0.988611199 0.986660352 0.984562067 0.982318271 0.979931013 0.977402455 0.974734872 0.971930643 0.968992248 0.965922263 0.962723352 0.959398265 0.955949832 0.952380952 0.948694596 0.944893794 0.940981632 0.936961247 0.932835821 0.928608573 0.924282757 0.919861653 0.915348565 0.910746812 0.906059727 0.901290648 0.896442915 0.891519863 0.886524823 0.88146111 0.876332025 0.871140846 0.865890828 0.860585198

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361

Appendix C Tables for Isentropic Flows (continued) γ = 1.4 Isentropic Temp Ratio Mach Number T/Tt

Isentropic Pressure Ratio p/pt

Isentropic Density Ratio ρ/ρt

Isentropic Area Ratio A/A*

Prandtl-Meyer Function v (degrees)

0.92 0.94 0.96 0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24 1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42 1.44 1.46 1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68 1.7 1.72 1.74 1.76 1.78 1.8

0.578475773 0.565775357 0.553169566 0.540668533 0.528281788 0.516018259 0.503886279 0.491893592 0.480047365 0.468354195 0.456820126 0.445450658 0.434250764 0.423224904 0.412377042 0.401710664 0.391228791 0.380934004 0.370828457 0.360913895 0.351191678 0.341662794 0.332327881 0.323187244 0.314240876 0.305488472 0.29692945 0.288562969 0.280387945 0.272403066 0.264606813 0.256997471 0.249573148 0.242331786 0.23527118 0.228388988 0.221682746 0.21514988 0.208787717 0.202593498 0.19656439 0.190697491 0.184989848 0.179438457 0.17404028

0.676400152 0.665759179 0.65512978 0.644520144 0.633938145 0.623391338 0.612886959 0.60243192 0.592032814 0.58169591 0.571427159 0.561232193 0.55111633 0.541084575 0.53114163 0.521291894 0.511539469 0.50188817 0.492341526 0.482902792 0.473574954 0.464360737 0.455262611 0.446282802 0.437423299 0.428685862 0.420072031 0.411583134 0.403220296 0.394984446 0.38687633 0.378896512 0.37104539 0.3633232 0.355730024 0.348265801 0.340930329 0.333723282 0.326644207 0.31969254 0.312867608 0.306168637 0.299594758 0.293145019 0.286818382

1.005597068 1.003108275 1.001364565 1.000337135 1 1.000329724 1.001305188 1.002907379 1.005119207 1.007925341 1.011312058 1.015267118 1.01977964 1.024840005 1.030439753 1.036571503 1.043228873 1.050406417 1.058099555 1.066304519 1.075018305 1.084238622 1.093963852 1.104193011 1.114925714 1.126162143 1.137903018 1.150149572 1.162903522 1.176167052 1.189942793 1.204233798 1.219043533 1.234375854 1.250235 1.266625574 1.283552533 1.30102118 1.319037148 1.337606397 1.3567352 1.376430139 1.396698095 1.417546247 1.438982058

− − − − 0 0.125688396 0.350982534 0.636686713 0.968039969 1.336200924 1.735039347 2.159960005 2.607346221 3.074255071 3.558233358 4.057198466 4.569357035 5.093147097 5.627195529 6.170285886 6.721333523 7.279365987 7.843507289 8.412965121 8.987020318 9.565018091 10.14636065 10.73050094 11.3169373 11.90520883 12.49489142 13.08559425 13.67695673 14.26864583 14.86035366 15.45179536 16.04270725 16.63284507 17.22198252 17.80990982 18.39643254 18.9813704 19.56455633 20.14583544 20.72506425

0.855227148 0.849819838 0.844366387 0.838869875 0.833333333 0.827759751 0.822152065 0.816513162 0.810845874 0.805152979 0.799437196 0.793701187 0.787947554 0.782178837 0.776397516 0.770606005 0.764806657 0.759001761 0.753193541 0.747384155 0.7415757 0.735770204 0.729969633 0.724175888 0.718390805 0.712616156 0.706853653 0.701104941 0.695371607 0.689655172 0.683957102 0.6782788 0.67262161 0.66698682 0.661375661 0.655789308 0.650228881 0.644695446 0.639190018 0.633713561 0.628266988 0.622851163 0.617466904 0.61211498 0.606796117

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362

Appendix C Tables for Isentropic Flows

(continued) γ = 1.4 Isentropic Temp Ratio Mach Number T/Tt

Isentropic Pressure Ratio p/pt

Isentropic Density Ratio ρ/ρt

Isentropic Area Ratio A/A*

Prandtl-Meyer Function v (degrees)

1.82 1.84 1.86 1.88 1.9 1.92 1.94 1.96 1.98 2 2.02 2.04 2.06 2.08 2.1 2.12 2.14 2.16 2.18 2.2 2.22 2.24 2.26 2.28 2.3 2.32 2.34 2.36 2.38 2.4 2.42 2.44 2.46 2.48 2.5 2.52 2.54 2.56 2.58 2.6 2.62 2.64 2.66 2.68 2.7

0.168792248 0.16369127 0.158734239 0.153918044 0.149239569 0.144695702 0.140283341 0.1359994 0.13184081 0.127804525 0.123887527 0.120086828 0.116399472 0.112822542 0.109353159 0.105988484 0.102725722 0.099562123 0.096494985 0.093521652 0.090639517 0.087846025 0.085138668 0.082514993 0.079972598 0.077509131 0.075122294 0.072809841 0.070569578 0.068399364 0.066297109 0.064260773 0.06228837 0.060377962 0.058527663 0.056735636 0.055000093 0.053319291 0.05169154 0.050115191 0.048588645 0.047110345 0.04567878 0.044292482 0.042950025

0.280613736 0.274529902 0.268565634 0.262719631 0.256990537 0.251376949 0.245877418 0.24049046 0.235214553 0.230048146 0.224989661 0.220037496 0.215190032 0.210445632 0.205802645 0.201259412 0.196814265 0.192465532 0.188211539 0.184050612 0.179981077 0.176001267 0.17210952 0.168304182 0.164583606 0.16094616 0.15739022 0.153914179 0.150516443 0.147195432 0.143949586 0.14077736 0.137677229 0.134647686 0.131687243 0.128794434 0.125967812 0.123205953 0.120507452 0.117870929 0.115295023 0.112778397 0.110319736 0.107917748 0.105571162

1.461013275 1.483647923 1.5068943 1.530760974 1.555256776 1.580390801 1.606172401 1.632611184 1.659717013 1.6875 1.715970508 1.745139146 1.775016769 1.805614476 1.836943609 1.869015753 1.901842733 1.935436615 1.969809704 2.004974545 2.040943923 2.077730858 2.115348612 2.153810685 2.193130815 2.233322979 2.274401391 2.316380509 2.359275026 2.403099877 2.447870237 2.493601525 2.540309398 2.588009758 2.63671875 2.686452762 2.737228428 2.789062629 2.841972489 2.895975385 2.951088938 3.007331021 3.06471976 3.123273528 3.183010955

21.30210986 21.87684925 22.44916864 23.01896282 23.58613467 24.15059458 24.71226003 25.27105507 25.82690997 26.37976081 26.92954915 27.47622164 28.01972978 28.56002957 29.0970813 29.63084922 30.16130138 30.68840936 31.21214807 31.73249557 32.24943286 32.76294373 33.27301459 33.77963431 34.28279408 34.78248726 35.2787093 35.77145756 36.26073121 36.74653115 37.22885988 37.70772142 38.1831212 38.65506599 39.12356383 39.5886239 40.05025652 40.50847302 40.9632857 41.41470778 41.86275331 42.30743714 42.74877486 43.18678274 43.62147769

0.601510996 0.596260256 0.591044494 0.585864267 0.580720093 0.575612452 0.570541786 0.565508505 0.560512981 0.555555556 0.550636536 0.5457562 0.540914795 0.536112541 0.531349628 0.526626222 0.521942461 0.517298461 0.512694311 0.508130081 0.503605818 0.499121546 0.494677273 0.490272984 0.485908649 0.481584219 0.47729963 0.473054799 0.468849631 0.464684015 0.460557828 0.456470932 0.452423179 0.448414407 0.444444444 0.44051311 0.43662021 0.432765545 0.428948904 0.425170068 0.421428812 0.417724903 0.414058101 0.410428159 0.406834825

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363

Appendix C Tables for Isentropic Flows (continued) γ = 1.4 Isentropic Temp Ratio Mach Number T/Tt

Isentropic Pressure Ratio p/pt

Isentropic Density Ratio ρ/ρt

Isentropic Area Ratio A/A*

Prandtl-Meyer Function v (degrees)

2.72 2.74 2.76 2.78 2.8 2.82 2.84 2.86 2.88 2.9 2.92 2.94 2.96 2.98 3 3.02 3.04 3.06 3.08 3.1 3.12 3.14 3.16 3.18 3.2 3.22 3.24 3.26 3.28 3.3 3.32 3.34 3.36 3.38 3.4 3.42 3.44 3.46 3.48 3.5 3.52 3.54 3.56 3.58 3.6

0.041650025 0.040391136 0.039172056 0.037991519 0.036848298 0.035741201 0.034669076 0.033630801 0.032625293 0.0316515 0.030708402 0.029795014 0.028910377 0.028053566 0.027223684 0.02641986 0.025641255 0.024887052 0.024156464 0.023448726 0.022763101 0.022098872 0.021455347 0.020831856 0.020227752 0.019642408 0.019075215 0.018525588 0.017992959 0.017476778 0.016976514 0.016491653 0.016021698 0.015566168 0.015124598 0.014696539 0.014281555 0.013879226 0.013489146 0.01311092 0.012744169 0.012388524 0.012043631 0.011709145 0.011384733

0.103278733 0.101039235 0.098851468 0.096714251 0.094626429 0.092586868 0.090594455 0.088648101 0.086746739 0.084889322 0.083074827 0.08130225 0.07957061 0.077878944 0.076226314 0.074611799 0.073034499 0.071493532 0.06998804 0.068517179 0.067080126 0.065676079 0.064304249 0.06296387 0.06165419 0.060374476 0.059124011 0.057902097 0.05670805 0.055541201 0.0544009 0.053286511 0.052197411 0.051132995 0.05009267 0.049075859 0.048081998 0.047110536 0.046160936 0.045232674 0.044325239 0.043438131 0.042570863 0.041722961 0.040893961

3.243950925 3.306112575 3.369515302 3.434178758 3.500122857 3.567367771 3.635933935 3.705842048 3.777113071 3.849768233 3.923829029 3.999317224 4.076254852 4.154664218 4.234567901 4.315988754 4.398949905 4.483474761 4.569587006 4.657310605 4.746669805 4.837689136 4.930393413 5.024807738 5.1209575 5.218868377 5.318566341 5.420077652 5.523428868 5.628646842 5.735758723 5.844791961 5.955774305 6.068733808 6.183698824 6.300698014 6.419760349 6.540915103 6.664191866 6.789620536 6.917231326 7.047054766 7.1791217 7.313463293 7.450111029

44.05287723 44.48099942 44.90586283 45.32748652 45.74588996 46.16109305 46.57311605 46.98197956 47.38770447 47.79031199 48.18982355 48.58626084 48.97964573 49.37000029 49.75734674 50.14170747 50.52310495 50.9015618 51.27710069 51.64974438 52.0195157 52.3864375 52.75053266 53.11182409 53.47033468 53.82608734 54.17910494 54.52941032 54.87702629 55.22197561 55.56428098 55.90396503 56.24105033 56.57555935 56.90751449 57.23693806 57.56385225 57.88827915 58.21024077 58.52975896 58.84685549 59.16155198 59.47386994 59.78383073 60.09145559

0.403277842 0.399756948 0.396271874 0.39282235 0.3894081 0.386028844 0.382684301 0.379374184 0.376098207 0.372856078 0.369647504 0.366472192 0.363329845 0.360220167 0.357142857 0.354097618 0.351084148 0.348102147 0.345151314 0.342231348 0.339341948 0.336482812 0.333653641 0.330854133 0.32808399 0.325342911 0.322630601 0.319946761 0.317291096 0.31466331 0.312063112 0.309490208 0.306944308 0.304425124 0.301932367 0.299465753 0.297024998 0.294609819 0.292219936 0.289855072 0.287514951 0.285199297 0.28290784 0.280640309 0.278396437

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364

Appendix C Tables for Isentropic Flows

(continued) γ = 1.4 Isentropic Temp Ratio Mach Number T/Tt

Isentropic Pressure Ratio p/pt

Isentropic Density Ratio ρ/ρt

Isentropic Area Ratio A/A*

Prandtl-Meyer Function v (degrees)

3.62 3.64 3.66 3.68 3.7 3.72 3.74 3.76 3.78 3.8 3.82 3.84 3.86 3.88 3.9 3.92 3.94 3.96 3.98 4 4.02 4.04 4.06 4.08 4.1 4.12 4.14 4.16 4.18 4.2 4.22 4.24 4.26 4.28 4.3 4.32 4.34 4.36 4.38 4.4 4.42 4.44 4.46 4.48 4.5

0.011070074 0.010764858 0.010468782 0.010181557 0.009902902 0.009632543 0.009370219 0.009115676 0.008868666 0.008628953 0.008396306 0.008170504 0.00795133 0.007738578 0.007532045 0.007331538 0.007136867 0.006947852 0.006764315 0.006586087 0.006413003 0.006244904 0.006081634 0.005923046 0.005768995 0.005619341 0.00547395 0.00533269 0.005195435 0.005062063 0.004932455 0.004806497 0.004684078 0.004565091 0.004449431 0.004336998 0.004227696 0.004121429 0.004018106 0.00391764 0.003819945 0.003724937 0.003632538 0.003542669 0.003455256

0.040083411 0.039290869 0.038515906 0.037758101 0.037017046 0.03629234 0.035583595 0.034890431 0.034212476 0.033549369 0.032900759 0.0322663 0.031645659 0.031038507 0.030444527 0.029863406 0.029294842 0.028738538 0.028194207 0.027661567 0.027140343 0.026630268 0.02613108 0.025642526 0.025164358 0.024696332 0.024238212 0.02378977 0.023350779 0.022921021 0.022500282 0.022088354 0.021685033 0.021290122 0.020903426 0.020524757 0.020153932 0.01979077 0.019435097 0.019086742 0.018745538 0.018411322 0.018083937 0.017763226 0.017449041

7.589096715 7.730452481 7.874210783 8.020404403 8.169066453 8.320230374 8.47392994 8.630199258 8.789072773 8.950585263 9.114771849 9.281667989 9.451309487 9.623732487 9.798973482 9.97706931 10.15805716 10.34197457 10.52885944 10.71875 10.91168486 11.10770299 11.30684369 11.50914665 11.71465192 11.9233999 12.13543136 12.35078745 12.56950968 12.79163993 13.01722046 13.24629391 13.47890327 13.71509194 13.95490369 14.19838267 14.4455734 14.69652081 14.9512702 15.20986727 15.47235811 15.73878919 16.00920738 16.28365995 16.56219457

60.39676563 60.6997818 61.00052491 61.29901565 61.59527452 61.88932191 62.18117804 62.47086297 62.75839662 63.04379875 63.32708895 63.60828668 63.88741122 64.16448169 64.43951705 64.71253612 64.98355752 65.25259975 65.51968113 65.7848198 66.04803376 66.30934086 66.56875876 66.82630496 67.08199684 67.33585156 67.58788616 67.83811751 68.08656232 68.33323714 68.57815837 68.82134223 69.06280481 69.30256204 69.54062968 69.77702335 70.01175851 70.24485046 70.47631437 70.70616525 70.93441794 71.16108717 71.3861875 71.60973333 71.83173895

0.276175957 0.273978608 0.271804127 0.269652256 0.267522739 0.265415322 0.263329752 0.26126578 0.25922316 0.257201646 0.255200996 0.253220971 0.251261332 0.249321845 0.247402276 0.245502396 0.243621977 0.241760792 0.23991862 0.238095238 0.236290429 0.234503977 0.232735668 0.230985291 0.229252636 0.227537498 0.225839672 0.224158956 0.22249515 0.220848057 0.219217481 0.217603231 0.216005115 0.214422945 0.212856535 0.2113057 0.20977026 0.208250033 0.206744844 0.205254516 0.203778875 0.202317752 0.200870977 0.199438382 0.198019802

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365

Appendix C Tables for Isentropic Flows (continued) γ = 1.4 Isentropic Temp Ratio Mach Number T/Tt

Isentropic Pressure Ratio p/pt

Isentropic Density Ratio ρ/ρt

Isentropic Area Ratio A/A*

Prandtl-Meyer Function v (degrees)

4.52 4.54 4.56 4.58 4.6 4.62 4.64 4.66 4.68 4.7 4.72 4.74 4.76 4.78 4.8 4.82 4.84 4.86 4.88 4.9 4.92 4.94 4.96 4.98 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 12 14 16 18 20

0.003370225 0.003287506 0.003207031 0.003128734 0.003052551 0.00297842 0.002906281 0.002836075 0.002767747 0.002701242 0.002636507 0.00257349 0.002512143 0.002452416 0.002394264 0.00233764 0.002282503 0.002228807 0.002176514 0.002125582 0.002075974 0.002027651 0.001980578 0.001934718 0.001890038 0.001074826 0.000633361 0.000385468 0.000241555 0.000155426 0.000102429 6.89843E-05 4.7386E-05 3.31411E-05 2.35631E-05 6.92218E-06 2.42778E-06 9.73091E-07 4.32722E-07 2.09075E-07

0.017141232 0.016839657 0.016544175 0.016254649 0.015970947 0.015692937 0.015420493 0.01515349 0.014891809 0.014635329 0.014383937 0.01413752 0.013895967 0.013659173 0.013427031 0.01319944 0.012976301 0.012757516 0.01254299 0.01233263 0.012126346 0.011924049 0.011725653 0.011531074 0.01134023 0.007577522 0.005193563 0.003642675 0.002608799 0.001903969 0.001413521 0.001065808 0.00081504 0.000631339 0.000494825 0.000206281 9.75966E-05 5.07953E-05 2.84731E-05 1.69351E-05

16.84485931 17.13170263 17.42277339 17.71812089 18.01779478 18.32184517 18.63032256 18.94327785 19.26076236 19.58282785 19.90952645 20.24091076 20.57703376 20.91794887 21.26370994 21.61437123 21.96998743 22.33061368 22.69630552 23.06711895 23.44311039 23.8243367 24.21085517 24.60272354 25 36.86896307 53.17978395 75.1343149 104.1428571 141.8414834 190.109375 251.0861673 327.1893004 421.1313734 535.9375 1276.214892 2685.383929 5144.554688 9159.28215 15377.34375

72.05221848 72.2711859 72.48865506 72.70463966 72.91915326 73.13220928 73.34382102 73.55400161 73.76276409 73.97012132 74.17608606 74.38067093 74.58388842 74.78575088 74.98627054 75.18545952 75.38332979 75.5798932 75.77516149 75.96914627 76.16185903 76.35331114 76.54351385 76.7324783 76.92021551 81.24479016 84.95549818 88.16816076 90.97273233 93.43966839 95.62467171 97.57220575 99.31809865 100.8914832 102.3162532 106.8786261 110.1798378 112.6758959 114.6278183 116.1952976

0.196615075 0.195224039 0.193846536 0.192482407 0.191131498 0.189793656 0.188468729 0.187156568 0.185857024 0.184569952 0.183295208 0.182032649 0.180782136 0.179543529 0.17831669 0.177101486 0.175897782 0.174705447 0.173524349 0.172354361 0.171195354 0.170047205 0.168909789 0.167782983 0.166666667 0.141843972 0.12195122 0.105820106 0.092592593 0.081632653 0.072463768 0.064724919 0.058139535 0.052493438 0.047619048 0.033557047 0.024875622 0.019157088 0.015197568 0.012345679

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366

Appendix C Tables for Isentropic Flows

γ = 5/3 Isentropic Temp Ratio Mach Number T/Tt

Isentropic Pressure Ratio p/pt

Isentropic Density Ratio ρ/ρt

Isentropic Area Ratio A/A*

Prandtl-Meyer Function v (degrees)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9

1 0.999666744 0.99866791 0.997006289 0.994686514 0.991715036 0.988100079 0.983851597 0.978981214 0.973502154 0.967429169 0.960778452 0.953567548 0.945815254 0.937541522 0.928767347 0.91951466 0.909806214 0.899665474 0.889116496 0.878183818 0.866892346 0.855267241 0.843333813 0.831117416 0.818643343 0.805936735 0.793022488 0.779925164 0.766668916 0.75327741 0.739773757 0.726180455 0.712519331 0.698811493 0.685077287 0.671336263 0.657607142 0.643907794 0.63025522 0.616665536 0.603153969 0.58973485 0.576421617 0.563226821 0.550162136

1 0.999800033 0.999200533 0.998202696 0.996808512 0.995020753 0.992842959 0.990279428 0.987335187 0.984015977 0.980328225 0.976279011 0.971876045 0.967127625 0.962042607 0.956630367 0.95090076 0.94486408 0.938531022 0.931912636 0.925020288 0.917865616 0.910460487 0.902816958 0.894947233 0.886863621 0.878578499 0.870104273 0.861453341 0.852638057 0.843670699 0.834563434 0.825328293 0.815977138 0.806521638 0.796973244 0.787343169 0.777642366 0.767881508 0.758070979 0.748220851 0.738340879 0.728440486 0.718528759 0.708614438 0.698705913

28.1325005 14.077504 9.3975135 7.061282 5.6625625 4.732608 4.070528643 3.575881 3.1928645 2.888 2.639983682 2.434614 2.262060038 2.115300571 1.9891875 1.8798605 1.784368265 1.700416 1.626192658 1.56025 1.501416214 1.448733091 1.401409587 1.358787 1.3203125 1.285518769 1.254008167 1.225440286 1.199522086 1.176 1.154653565 1.13529025 1.117741227 1.101857882 1.087508929 1.074578 1.062961635 1.052567579 1.043313346 1.035125 1.02793611 1.021686857 1.016323267 1.011796545 1.0080625

− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −

1 0.999866684 0.999466951 0.998801438 0.997871208 0.996677741 0.99522293 0.993509074 0.991538868 0.989315394 0.986842105 0.984122819 0.981161695 0.977963229 0.974532225 0.970873786 0.966993296 0.962896392 0.958588957 0.954077089 0.949367089 0.944465433 0.939378758 0.934113837 0.928677563 0.923076923 0.917318982 0.911410864 0.90535973 0.899172761 0.892857143 0.886420045 0.879868606 0.87320992 0.866451017 0.859598854 0.8526603 0.845642124 0.838550984 0.831393415 0.824175824 0.816904477 0.809585492 0.802224837 0.794828317 0.787401575

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367

Appendix C Tables for Isentropic Flows (continued) γ = 5/3 Isentropic Temp Ratio Mach Number T/Tt

Isentropic Pressure Ratio p/pt

Isentropic Density Ratio ρ/ρt

Isentropic Area Ratio A/A*

Prandtl-Meyer Function v (degrees)

0.92 0.94 0.96 0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24 1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42 1.44 1.46 1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68 1.7 1.72 1.74 1.76 1.78 1.8

0.537238369 0.524465477 0.511852586 0.499408014 0.48713929 0.475053181 0.463155722 0.451452237 0.439947375 0.428645133 0.417548893 0.406661447 0.395985028 0.385521346 0.375271611 0.365236567 0.355416523 0.345811377 0.336420647 0.327243501 0.318278778 0.309525018 0.300980486 0.292643193 0.284510924 0.276581252 0.268851569 0.261319095 0.253980905 0.246833942 0.239875034 0.233100913 0.226508225 0.220093547 0.213853399 0.207784256 0.201882558 0.196144722 0.190567151 0.185146241 0.179878393 0.174760016 0.169787537 0.164957405 0.160266098

0.688811221 0.678938042 0.669093701 0.659285166 0.649519053 0.639801624 0.630138798 0.620536148 0.610998914 0.601532004 0.592140004 0.582827185 0.573597513 0.564454653 0.555401984 0.546442603 0.537579338 0.528814757 0.520151176 0.511590673 0.503135092 0.494786059 0.486544988 0.478413093 0.470391394 0.462480732 0.454681773 0.446995023 0.43942083 0.431959398 0.424610793 0.417374954 0.410251697 0.403240724 0.396341633 0.389553923 0.382877001 0.376310188 0.369852727 0.363503787 0.357262473 0.351127825 0.345098828 0.339174419 0.333353483

1.005081043 1.002815755 1.0012335 1.000304092 1 1.000296088 1.001169385 1.002598877 1.004565333 1.007051136 1.010040143 1.013517553 1.017469793 1.021884415 1.02675 1.032056074 1.037793032 1.043952071 1.050525125 1.057504808 1.064884364 1.072657619 1.080818941 1.089363196 1.098285714 1.107582261 1.117249 1.127282473 1.137679568 1.1484375 1.159553789 1.17102624 1.182852923 1.195032158 1.2075625 1.220442722 1.233671805 1.247248922 1.261173429 1.275444853 1.290062884 1.305027362 1.320338273 1.335995736 1.352

− − − − 0 0.112893887 0.314627205 0.569606709 0.864339518 1.190715632 1.543101358 1.917270958 2.309899831 2.71828616 3.140182376 3.573685743 4.017163459 4.469199208 4.928553752 5.394135038 5.864975027 6.340211359 6.819072621 7.300866322 7.784968967 8.270817765 8.757903635 9.245765258 9.733983977 10.22217939 10.71000553 11.19714752 11.68331861 12.16825763 12.65172665 13.13350895 13.6134072 14.09124177 14.56684934 15.04008148 15.51080351 15.97889337 16.44424066 16.90674571 17.36631874

0.779950083 0.772479143 0.76499388 0.757499243 0.75 0.742500743 0.73500588 0.727519643 0.720046083 0.712589074 0.705152313 0.697739325 0.690353461 0.682997905 0.675675676 0.668389627 0.661142454 0.653936699 0.64677475 0.639658849 0.632591093 0.625573442 0.61860772 0.61169562 0.60483871 0.598038434 0.591296121 0.584612986 0.577990136 0.571428571 0.564929196 0.558492814 0.552120141 0.545811804 0.539568345 0.533390228 0.52727784 0.521231496 0.515251443 0.509337861 0.50349087 0.497710532 0.491996851 0.486349783 0.480769231

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368

Appendix C Tables for Isentropic Flows

(continued) γ = 5/3 Isentropic Temp Ratio Mach Number T/Tt

Isentropic Pressure Ratio p/pt

Isentropic Density Ratio ρ/ρt

Isentropic Area Ratio A/A*

Prandtl-Meyer Function v (degrees)

1.82 1.84 1.86 1.88 1.9 1.92 1.94 1.96 1.98 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 12 14 16 18 20

0.155710127 0.151286042 0.146990434 0.14281994 0.138771246 0.134841089 0.131026261 0.127323609 0.123730037 0.120242511 0.090575967 0.068634563 0.05238158 0.04029303 0.03125 0.024438986 0.019270798 0.015318665 0.01227261 0.009906475 0.005980614 0.003757578 0.002445254 0.001641125 0.001131766 0.000799456 0.000576876 0.000424244 0.000317349 0.000241049 0.000185644 0.000144781 5.9499E-05 2.79043E-05 1.44396E-05 8.06183E-06 4.78124E-06

0.327634869 0.322017383 0.316499802 0.311080872 0.305759312 0.30053382 0.295403073 0.290365733 0.28542045 0.280565859 0.236705194 0.200412923 0.17041474 0.14559215 0.125 0.107857389 0.093527606 0.081495298 0.071344773 0.062741006 0.046349755 0.035070732 0.027101567 0.021334623 0.017070805 0.013857244 0.011393295 0.009474792 0.007960166 0.006749366 0.005770436 0.004970797 0.002915452 0.001850982 0.001246615 0.00087874 0.00064228

1.368351434 1.385050522 1.402097855 1.419494128 1.437240132 1.45533675 1.473784954 1.492585796 1.511740409 1.53125 1.746181818 1.998375 2.289846154 2.622892857 3 3.42378125 3.896941176 4.42225 5.002526316 5.640625 7.5078125 9.8 12.56321023 15.84375 19.68810096 24.14285714 29.2546875 35.0703125 41.63648897 49 57.20764803 66.30625 112.546875 176.7901786 262.0351563 371.28125 507.528125

17.82287917 18.27635486 18.7266815 19.17380205 19.61766618 20.05822975 20.49545443 20.92930719 21.35975997 21.7867893 25.86630898 29.60144219 33.00872276 36.1131625 38.94244127 41.52398701 43.88357677 46.04474819 48.02863993 49.85405234 53.82857743 57.12165044 59.886881 62.23721625 64.25675012 66.0089832 67.54255199 68.89522082 70.09669137 71.17060491 72.13598787 73.00830955 75.79069327 77.79475833 79.30566282 80.48490123 81.43058911

0.475255054 0.469807066 0.464425042 0.459108717 0.453857791 0.448671931 0.443550772 0.43849392 0.433500954 0.428571429 0.382653061 0.342465753 0.307377049 0.276752768 0.25 0.226586103 0.206043956 0.187969925 0.172018349 0.157894737 0.129032258 0.107142857 0.090225564 0.076923077 0.066298343 0.057692308 0.050632911 0.044776119 0.03986711 0.035714286 0.032171582 0.029126214 0.020408163 0.015075377 0.011583012 0.009174312 0.007444169

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369

Appendix C Tables for Isentropic Flows

γ = 1.3 Isentropic Temp Ratio Mach Number T/Tt

Isentropic Pressure Ratio p/pt

Isentropic Density Ratio ρ/ρt

Isentropic Area Ratio A/A*

Prandtl-Meyer Function v (degrees)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9

1 0.999740042 0.998960665 0.997663366 0.995850628 0.993525918 0.990693669 0.987359265 0.983529022 0.979210162 0.974410788 0.969139853 0.963407129 0.957223168 0.950599269 0.943547432 0.93608032 0.928211214 0.919953966 0.911322954 0.902333029 0.892999474 0.883337948 0.873364438 0.863095212 0.852546763 0.841735768 0.83067903 0.819393439 0.807895919 0.796203385 0.784332697 0.772300621 0.760123783 0.747818632 0.735401405 0.722888088 0.710294384 0.697635684 0.684927035 0.672183115 0.659418209 0.646646188 0.633880486 0.621134087 0.608419506

1 0.999800026 0.999200416 0.998202104 0.996806645 0.995016207 0.992833567 0.990262101 0.987305773 0.983969123 0.980257253 0.976175809 0.971730967 0.966929411 0.961778316 0.956285322 0.950458514 0.944306397 0.937837872 0.931062209 0.923989022 0.91662824 0.908990082 0.901085026 0.892923782 0.884517267 0.87587657 0.867012931 0.857937707 0.848662347 0.839198368 0.829557321 0.819750771 0.809790271 0.799687332 0.789453408 0.779099866 0.768637965 0.75807884 0.747433476 0.736712694 0.72592713 0.71508722 0.704203187 0.693285023 0.682342476

29.2681205 14.64415972 9.774002412 7.342304564 5.886000133 4.917402747 4.227505951 3.711808266 3.312254883 2.994014338 2.734922458 2.520204335 2.339632315 2.185901989 2.053657791 1.938884001 1.838510679 1.750150971 1.671921354 1.602315822 1.540116015 1.484325856 1.434123245 1.388823828 1.347853461 1.310727031 1.277031984 1.24641538 1.218573635 1.193244304 1.170199466 1.149240341 1.130192893 1.112904206 1.09723948 1.083079535 1.070318715 1.058863121 1.048629125 1.039542101 1.031535341 1.024549133 1.018529963 1.013429827 1.009205638

– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

1 0.999940004 0.999760058 0.999460291 0.999040921 0.998502247 0.997844656 0.997068618 0.996174689 0.995163505 0.994035785 0.992792328 0.99143401 0.989961787 0.98837669 0.986679822 0.984872361 0.982955551 0.980930707 0.978799209 0.9765625 0.974222084 0.971779523 0.969236436 0.966594494 0.963855422 0.961020989 0.958093012 0.95507335 0.951963902 0.948766603 0.945483426 0.94211637 0.938667468 0.935138775 0.931532371 0.927850356 0.924094849 0.920267982 0.9163719 0.912408759 0.908380721 0.904289952 0.900138621 0.895928899 0.891662951

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370

Appendix C Tables for Isentropic Flows

(continued) γ = 1.3 Isentropic Temp Ratio Mach Number T/Tt

Isentropic Pressure Ratio p/pt

Isentropic Density Ratio ρ/ρt

Isentropic Area Ratio A/A*

Prandtl-Meyer Function v (degrees)

0.92 0.94 0.96 0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24 1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42 1.44 1.46 1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68 1.7 1.72 1.74 1.76 1.78 1.8

0.595748777 0.583133446 0.570584555 0.558112642 0.545727734 0.533439342 0.521256465 0.509187585 0.497240673 0.485423191 0.4737421 0.462203862 0.450814454 0.439579369 0.428503633 0.417591813 0.406848029 0.396275964 0.385878882 0.375659637 0.36562069 0.355764121 0.346091648 0.336604637 0.327304118 0.318190804 0.309265102 0.300527132 0.291976738 0.283613505 0.275436776 0.267445662 0.259639061 0.252015669 0.244573994 0.237312372 0.230228976 0.223321832 0.216588829 0.210027733 0.203636197 0.19741177 0.191351912 0.185454002 0.179715344

0.671385042 0.660421953 0.649462164 0.638514349 0.627586894 0.616687886 0.605825114 0.595006061 0.584237901 0.5735275 0.562881413 0.552305883 0.541806843 0.531389916 0.521060418 0.510823362 0.500683458 0.490645122 0.480712476 0.470889354 0.461179313 0.45158563 0.442111315 0.432759117 0.423531528 0.414430794 0.40545892 0.396617677 0.387908615 0.379333063 0.370892145 0.362586782 0.354417704 0.346385456 0.338490408 0.33073276 0.323112554 0.315629678 0.308283876 0.301074756 0.294001795 0.287064351 0.280261665 0.273592871 0.267057002

1.005818704 1.003234285 1.001421195 1.000351457 1 1.000344396 1.001364619 1.00304284 1.005363245 1.008311862 1.011876426 1.016046242 1.020812072 1.026166029 1.032101486 1.038612993 1.045696202 1.053347799 1.061565445 1.070347724 1.079694088 1.089604822 1.100080996 1.111124436 1.122737691 1.134924 1.147687272 1.161032061 1.174963544 1.189487504 1.204610313 1.220338919 1.236680829 1.253644103 1.271237342 1.289469675 1.308350758 1.327890765 1.348100382 1.368990802 1.390573724 1.412861348 1.435866375 1.459602002 1.484081927

– – – – 0 0.131267258 0.36688068 0.666106121 1.013653043 1.40038103 1.819961129 2.267653326 2.73972878 3.233153684 3.745398608 4.274315179 4.818052017 5.374994979 5.943723273 6.522976317 7.111628141 7.708667227 8.313180366 8.924339548 9.541391177 10.16364709 10.79047705 11.42130231 12.05559022 12.69284948 13.33262613 13.97449997 14.61808149 15.26300913 15.90894688 16.55558215 17.20262382 17.84980059 18.49685942 19.14356415 19.78969427 20.43504378 21.07942014 21.72264341 22.36454532

0.88734294 0.882971021 0.878549339 0.874080031 0.869565217 0.865007007 0.860407489 0.855768737 0.851092803 0.846381718 0.84163749 0.836862102 0.832057512 0.827225651 0.822368421 0.817487697 0.812585321 0.807663108 0.802722836 0.797766254 0.792795078 0.787810988 0.782815631 0.777810619 0.772797527 0.767777897 0.762753234 0.757725006 0.752694647 0.747663551 0.74263308 0.737604555 0.732579265 0.727558459 0.722543353 0.717535123 0.712534914 0.707543832 0.70256295 0.697593303 0.692635895 0.687691694 0.682761634 0.677846617 0.67294751

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371

Appendix C Tables for Isentropic Flows (continued) γ = 1.3 Isentropic Temp Ratio Mach Number T/Tt

Isentropic Pressure Ratio p/pt

Isentropic Density Ratio ρ/ρt

Isentropic Area Ratio A/A*

Prandtl-Meyer Function v (degrees)

1.82 1.84 1.86 1.88 1.9 1.92 1.94 1.96 1.98 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 12 14 16 18 20

0.174133184 0.168704709 0.163427066 0.15829736 0.153312668 0.148470042 0.143766519 0.139199126 0.134764882 0.130460811 0.093934091 0.067308162 0.048129237 0.034420048 0.024662623 0.017729096 0.012799896 0.009288299 0.006778278 0.004976508 0.002363265 0.001168587 0.000601137 0.000321036 0.000177538 0.000101397 5.96513E-05 3.60585E-05 2.23464E-05 1.41684E-05 9.1737E-06 6.05545E-06 1.3558E-06 3.75158E-07 1.2195E-07 4.49689E-08 1.83472E-08

0.260652997 0.254379709 0.248235908 0.242220289 0.236331477 0.230568037 0.22492847 0.21941123 0.214014719 0.208737298 0.162130241 0.125462414 0.096932284 0.074898025 0.057957164 0.044960987 0.034994917 0.027344752 0.021460029 0.016920127 0.009541681 0.005550788 0.003328799 0.002054629 0.001302686 0.000846665 0.000562959 0.00038222 0.000264525 0.000186315 0.000133363 9.68873E-05 3.06412E-05 1.14048E-05 4.80484E-06 2.23046E-06 1.11918E-06

1.509320343 1.535331944 1.562131923 1.589735972 1.618160289 1.647421578 1.677537051 1.708524433 1.740401964 1.773188407 2.155552632 2.653523986 3.295436021 4.116479872 5.159771816 6.477594472 8.132804216 10.20040384 12.76928281 15.94412912 27.38695934 45.9565175 75.21969871 120.0964754 187.2173211 285.3371996 425.8095325 623.1234688 895.5076955 1265.60395 1761.213327 2416.118397 7566.459573 20209.03829 47783.04401 102659.4343 204201.683

23.00496853 23.6437659 24.28079986 24.91594177 25.5490714 26.18007636 26.80885169 27.43529939 28.05932797 28.68085215 34.74334953 40.49622775 45.91676103 51.00152922 55.7584169 60.20174997 64.34926578 68.22020898 71.83414037 75.21020605 82.72711082 89.12342651 94.60781975 99.34724887 103.4742577 107.0940706 110.2905606 113.1309908 115.6696866 117.9508535 120.0107342 121.8792625 127.8816248 132.2409423 135.5448414 138.1326204 140.2130595

0.66806515 0.66320034 0.658353852 0.653526429 0.64871878 0.643931589 0.639165506 0.634421154 0.62969913 0.625 0.579374276 0.536480687 0.49652433 0.459558824 0.425531915 0.394321767 0.365764448 0.339673913 0.31585597 0.294117647 0.247678019 0.210526316 0.180586907 0.15625 0.136286201 0.119760479 0.105960265 0.094339623 0.084477297 0.076045627 0.068787618 0.0625 0.044247788 0.032894737 0.025380711 0.02016129 0.016393443

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APPENDIX D

Tables for Normal Shock Waves

Note – The numerical values of the table can be directly obtained from the following equations. They are tabulated for γ ¼ 1:4 for convenience. They are applicable for oblique shock waves whose normal and tangential components of Mach number are denoted, respectively, by Mn and Mt . For normal shock waves, M1n ¼ M1 and M2n ¼ M2 . (The subscripts t for thermodynamic properties denote stagnation values; the subscripts 1 and 2 denote upstream and downstream of the shock, respectively. The superscript * denotes critical condition for M ¼ 1.) 2 ¼ M2n

2 2 M1n þ γ1 2γ 2 γ1 M1n

1

" #γ=ðγ1Þ " #1=ðγ1Þ γþ1 2 pt2 A1 1 2 M1n ¼ ¼ 2γ γ1 2 2 pt1 A2 1 þ γ1 2 M1n γþ1 M1n  γþ1 2 p2 2γM1n γ1  ¼ p1 γþ1 γþ1

T2 ¼ T1



M1t M2t

2

 
2γ 2 2 1 þ γ1 M M  1 1n 1n 2 γ1 h i ¼ ðγþ1Þ2 2 2ðγ1Þ M1n

(4.22a)

(4.22b)

(4.22c)




(4.22d)

pffiffiffiffiffiffiffiffiffiffiffi 2 ðγ þ 1ÞM1n ρ2 V1n M1n γRT1 pffiffiffiffiffiffiffiffiffiffiffi ¼ ¼ ¼ 2 þ2 ρ1 V2n M2n γRT2 ðγ  1ÞM1n

(4.22e)

Tt2 ¼1 Tt1

(4.22f)

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Appendix D Tables for Normal Shock Waves

373

γ = 1.4 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24 1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42 1.44 1.46 1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68 1.7 1.72 1.74 1.76 1.78 1.8 1.82 1.84 1.86 1.88 1.9 1.92 1.94

1 0.980519493 0.962025499 0.944445301 0.927713359 0.911770421 0.896562769 0.882041571 0.868162321 0.854884363 0.84217047 0.829986484 0.818300995 0.807085068 0.796311994 0.78595708 0.775997454 0.7664119 0.757180705 0.748285526 0.739709275 0.731436004 0.723450815 0.715739774 0.708289828 0.701088742 0.694125028 0.687387893 0.680867185 0.674553345 0.668437365 0.662510745 0.656765462 0.651193934 0.645788989 0.640543841 0.635452058 0.630507546 0.625704523 0.621037498 0.616501258 0.612090843 0.607801539 0.603628856 0.59956852 0.595616455 0.591768777 0.58802178

1 1.000009968 1.000076723 1.000249345 1.000569625 1.001073135 1.001790146 1.002746407 1.003963821 1.005461017 1.007253842 1.009355783 1.011778331 1.014531287 1.017623034 1.021060758 1.024850648 1.028998065 1.033507686 1.038383625 1.043629544 1.049248741 1.055244231 1.061618814 1.068375129 1.075515709 1.08304302 1.090959498 1.099267581 1.107969738 1.117068486 1.126566416 1.136466204 1.146770629 1.157482582 1.16860508 1.18014127 1.192094437 1.204468015 1.217265582 1.230490875 1.244147786 1.258240365 1.272772826 1.287749546 1.303175066 1.319054094 1.335391502

1 1.047133333 1.0952 1.1442 1.194133333 1.245 1.2968 1.349533333 1.4032 1.4578 1.513333333 1.5698 1.6272 1.685533333 1.7448 1.805 1.866133333 1.9282 1.9912 2.055133333 2.12 2.1858 2.252533333 2.3202 2.3888 2.458333333 2.5288 2.6002 2.672533333 2.7458 2.82 2.895133333 2.9712 3.0482 3.126133333 3.205 3.2848 3.365533333 3.4472 3.5298 3.613333333 3.6978 3.7832 3.869533333 3.9568 4.045 4.134133333 4.2242

1 1.01324878 1.02634497 1.039311606 1.052169608 1.064938017 1.077634184 1.090273954 1.102871819 1.115441051 1.127993827 1.140541333 1.153093861 1.165660891 1.178251172 1.190872781 1.20353319 1.216239318 1.228997578 1.241813921 1.254693878 1.267642591 1.280664849 1.293765115 1.306947553 1.320216049 1.333574238 1.347025519 1.360573073 1.374219885 1.38796875 1.401822295 1.415782986 1.429853143 1.444034946 1.45833045 1.47274159 1.487270191 1.501917975 1.516686567 1.531577503 1.546592235 1.561732136 1.576998506 1.592392576 1.607915512 1.623568422 1.639352354

1 1.033441494 1.067087609 1.100921027 1.134924754 1.169082126 1.203376823 1.237792876 1.272314675 1.306926976 1.341614907 1.376363973 1.411160059 1.445989435 1.480838756 1.515695067 1.5505458 1.585378775 1.6201822 1.654944673 1.689655172 1.724303061 1.758878082 1.793370352 1.82777036 1.862068966 1.896257387 1.930327202 1.96427034 1.998079078 2.031746032 2.065264152 2.098626717 2.131827325 2.16485989 2.197718631 2.23039807 2.262893019 2.295198577 2.327310122 2.359223301 2.390934026 2.422438466 2.453733037 2.484814398 2.515679443 2.54632529 2.576749281

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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374

Appendix D Tables for Normal Shock Waves

(continued) γ = 1.4 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

1.96 1.98 2 2.02 2.04 2.06 2.08 2.1 2.12 2.14 2.16 2.18 2.2 2.22 2.24 2.26 2.28 2.3 2.32 2.34 2.36 2.38 2.4 2.42 2.44 2.46 2.48 2.5 2.52 2.54 2.56 2.58 2.6 2.62 2.64 2.66 2.68 2.7 2.72 2.74 2.76 2.78 2.8 2.82 2.84 2.86 2.88

0.584371924 0.58081583 0.577350269 0.573972154 0.57067853 0.567466572 0.564333573 0.561276941 0.558294192 0.555382944 0.552540913 0.549765907 0.547055823 0.544408639 0.541822415 0.539295285 0.536825457 0.534411206 0.532050873 0.52974286 0.527485632 0.525277707 0.523117659 0.521004115 0.518935749 0.516911285 0.51492949 0.512989176 0.511089196 0.509228441 0.507405844 0.505620372 0.503871026 0.502156842 0.500476888 0.498830263 0.497216096 0.495633542 0.494081788 0.492560042 0.491067542 0.489603546 0.488167338 0.486758224 0.485375531 0.484018608 0.482686823

1.35219233 1.369461786 1.387205243 1.405428241 1.424136488 1.443335859 1.463032392 1.483232295 1.503941939 1.52516786 1.546916761 1.569195506 1.592011126 1.615370814 1.639281927 1.663751984 1.688788667 1.71439982 1.74059345 1.767377725 1.794760973 1.822751686 1.851358517 1.880590278 1.910455944 1.94096465 1.972125694 2.003948532 2.036442784 2.069618229 2.10348481 2.138052629 2.173331951 2.209333203 2.246066974 2.283544015 2.321775241 2.36077173 2.400544722 2.441105623 2.482466001 2.524637591 2.567632292 2.611462169 2.656139453 2.701676542 2.748086

4.3152 4.407133333 4.5 4.5938 4.688533333 4.7842 4.8808 4.978333333 5.0768 5.1762 5.276533333 5.3778 5.48 5.583133333 5.6872 5.7922 5.898133333 6.005 6.1128 6.221533333 6.3312 6.4418 6.553333333 6.6658 6.7792 6.893533333 7.0088 7.125 7.242133333 7.3602 7.4792 7.599133333 7.72 7.8418 7.964533333 8.0882 8.2128 8.338333333 8.4648 8.5922 8.720533333 8.8498 8.98 9.111133333 9.2432 9.3762 9.510133333

1.655268305 1.67131722 1.6875 1.703817498 1.720270528 1.736859864 1.753586243 1.770450365 1.787452901 1.804594489 1.821875735 1.839297222 1.856859504 1.87456311 1.892408546 1.910396296 1.928526822 1.946800567 1.965217955 1.983779391 2.002485263 2.021335944 2.04033179 2.059473144 2.078760333 2.098193673 2.117773465 2.1375 2.157373556 2.177394401 2.197562793 2.217878978 2.238343195 2.258955673 2.279716631 2.300626282 2.32168483 2.342892471 2.364249394 2.385755783 2.407411814 2.429217654 2.451173469 2.473279416 2.495535648 2.51794231 2.540499546

2.606948969 2.636922111 2.666666667 2.696180785 2.725462801 2.754511229 2.783324756 2.811902232 2.840242669 2.868345234 2.896209237 2.923834133 2.951219512 2.978365094 3.005270724 3.031936365 3.058362096 3.084548105 3.110494683 3.136202222 3.161671208 3.186902217 3.211895911 3.236653034 3.261174408 3.285460929 3.30951356 3.333333333 3.356921342 3.380278738 3.40340673 3.426306578 3.448979592 3.471427127 3.493650581 3.515651396 3.537431048 3.55899105 3.580332946 3.601458313 3.622368755 3.6430659 3.663551402 3.683826936 3.703894195 3.723754894 3.743410759

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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Appendix D Tables for Normal Shock Waves

375

(continued) γ = 1.4 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

2.9 2.92 2.94 2.96 2.98 3 3.02 3.04 3.06 3.08 3.1 3.12 3.14 3.16 3.18 3.2 3.22 3.24 3.26 3.28 3.3 3.32 3.34 3.36 3.38 3.4 3.42 3.44 3.46 3.48 3.5 3.52 3.54 3.56 3.58 3.6 3.62 3.64 3.66 3.68 3.7 3.72 3.74 3.76 3.78 3.8 3.82

0.481379562 0.480096233 0.478836259 0.477599082 0.476384159 0.475190963 0.474018985 0.472867729 0.471736712 0.470625468 0.469533542 0.468460494 0.467405894 0.466369326 0.465350385 0.464348678 0.463363821 0.462395442 0.46144318 0.460506682 0.459585605 0.458679617 0.457788392 0.456911616 0.456048979 0.455200184 0.454364938 0.453542957 0.452733966 0.451937694 0.45115388 0.450382266 0.449622605 0.448874652 0.448138172 0.447412933 0.44669871 0.445995283 0.445302438 0.444619967 0.443947664 0.443285332 0.442632777 0.441989808 0.441356241 0.440731896 0.440116596

2.795380561 2.843573126 2.892676765 2.942704717 2.993670393 3.045587373 3.09846941 3.152330428 3.207184524 3.263045968 3.319929207 3.37784886 3.436819723 3.496856768 3.557975145 3.620190181 3.683517382 3.747972434 3.813571204 3.880329738 3.948264267 4.017391202 4.087727139 4.159288859 4.232093327 4.306157695 4.381499303 4.458135677 4.536084532 4.615363776 4.695991502 4.777986 4.861365749 4.946149422 5.032355887 5.120004206 5.209113639 5.299703639 5.391793863 5.48540416 5.580554585 5.677265389 5.775557028 5.875450158 5.976965641 6.080124541 6.18494813

9.645 9.7808 9.917533333 10.0552 10.1938 10.33333333 10.4738 10.6152 10.75753333 10.9008 11.045 11.19013333 11.3362 11.4832 11.63113333 11.78 11.9298 12.08053333 12.2322 12.3848 12.53833333 12.6928 12.8482 13.00453333 13.1618 13.32 13.47913333 13.6392 13.8002 13.96213333 14.125 14.2888 14.45353333 14.6192 14.7858 14.95333333 15.1218 15.2912 15.46153333 15.6328 15.805 15.97813333 16.1522 16.3272 16.50313333 16.68 16.8578

2.563207491 2.586066279 2.609076037 2.632236888 2.655548953 2.679012346 2.702627179 2.72639356 2.750311593 2.77438138 2.798603018 2.822976602 2.847502223 2.872179971 2.897009931 2.921992188 2.94712682 2.972413907 2.997853525 3.023445747 3.049190644 3.075088286 3.101138739 3.12734207 3.15369834 3.180207612 3.206869946 3.233685398 3.260654025 3.287775881 3.31505102 3.342479494 3.370061351 3.397796642 3.425685412 3.453727709 3.481923577 3.510273059 3.538776197 3.567433034 3.596243608 3.62520796 3.654326126 3.683598144 3.71302405 3.742603878 3.772337663

3.762863535 3.782114975 3.801166847 3.820020928 3.838679001 3.857142857 3.875414294 3.893495113 3.911387117 3.929092114 3.94661191 3.963948311 3.981103125 3.998078155 4.014875202 4.031496063 4.047942531 4.064216395 4.080319435 4.096253427 4.112020138 4.12762133 4.143058754 4.158334152 4.173449258 4.188405797 4.203205481 4.217850014 4.232341087 4.246680382 4.260869565 4.274910295 4.288804216 4.30255296 4.316158146 4.329621381 4.342944257 4.356128353 4.369175238 4.382086461 4.394863563 4.407508069 4.420021488 4.432405317 4.44466104 4.456790123 4.468794022

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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376

Appendix D Tables for Normal Shock Waves

(continued) γ = 1.4 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

3.84 3.86 3.88 3.9 3.92 3.94 3.96 3.98 4 4.02 4.04 4.06 4.08 4.1 4.12 4.14 4.16 4.18 4.2 4.22 4.24 4.26 4.28 4.3 4.32 4.34 4.36 4.38 4.4 4.42 4.44 4.46 4.48 4.5 4.52 4.54 4.56 4.58 4.6 4.62 4.64 4.66 4.68 4.7 4.72 4.74 4.76

0.43951017 0.438912449 0.438323269 0.43774247 0.437169895 0.436605392 0.436048812 0.435500007 0.434958836 0.434425159 0.43389884 0.433379746 0.432867746 0.432362714 0.431864523 0.431373054 0.430888186 0.430409804 0.429937793 0.429472041 0.429012441 0.428558884 0.428111268 0.427669488 0.427233447 0.426803044 0.426378186 0.425958777 0.425544726 0.425135944 0.424732341 0.424333832 0.423940332 0.423551759 0.423168031 0.422789069 0.422414795 0.422045133 0.421680008 0.421319347 0.420963079 0.420611132 0.420263438 0.41991993 0.419580541 0.419245205 0.418913861

6.291457885 6.399675491 6.50962284 6.621322036 6.734795392 6.850065431 6.96715489 7.08608672 7.206884085 7.329570363 7.454169151 7.580704263 7.709199728 7.839679798 7.972168945 8.106691859 8.243273455 8.381938872 8.522713471 8.66562284 8.810692791 8.957949367 9.107418836 9.259127698 9.413102681 9.569370745 9.727959085 9.888895125 10.05220653 10.21792119 10.38606724 10.55667305 10.72976723 10.90537863 11.08353634 11.26426968 11.44760824 11.63358182 11.8222205 12.01355458 12.2076146 12.40443138 12.60403597 12.80645967 13.01173403 13.21989084 13.43096218

17.03653333 17.2162 17.3968 17.57833333 17.7608 17.9442 18.12853333 18.3138 18.5 18.68713333 18.8752 19.0642 19.25413333 19.445 19.6368 19.82953333 20.0232 20.2178 20.41333333 20.6098 20.8072 21.00553333 21.2048 21.405 21.60613333 21.8082 22.0112 22.21513333 22.42 22.6258 22.83253333 23.0402 23.2488 23.45833333 23.6688 23.8802 24.09253333 24.3058 24.52 24.73513333 24.9512 25.1682 25.38613333 25.605 25.8248 26.04553333 26.2672

3.802225439 3.832267237 3.862463089 3.892813025 3.923317076 3.953975271 3.984787638 4.015754206 4.046875 4.078150048 4.109579375 4.141163006 4.172900965 4.204793278 4.236839966 4.269041053 4.301396561 4.333906511 4.366570925 4.399389823 4.432363225 4.465491152 4.498773622 4.532210654 4.565802267 4.599548478 4.633449306 4.667504766 4.701714876 4.736079652 4.770599111 4.805273267 4.840102136 4.875085734 4.910224074 4.945517171 4.980965039 5.016567691 5.052325142 5.088237403 5.124304489 5.160526411 5.196903181 5.233434812 5.270121316 5.306962703 5.343958986

4.480674176 4.492432009 4.504068933 4.515586343 4.526985623 4.53826814 4.549435247 4.560488282 4.571428571 4.582257424 4.592976137 4.603585991 4.614088255 4.624484182 4.634775011 4.644961969 4.655046266 4.665029102 4.674911661 4.684695112 4.694380614 4.70396931 4.71346233 4.722860792 4.732165799 4.741378442 4.7504998 4.759530937 4.768472906 4.777326747 4.786093487 4.794774141 4.803369711 4.811881188 4.820309551 4.828655765 4.836920787 4.845105557 4.853211009 4.861238062 4.869187624 4.877060594 4.884857857 4.892580288 4.900228752 4.907804104 4.915307185

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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Appendix D Tables for Normal Shock Waves

377

(continued) γ = 1.4 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

4.78 4.8 4.82 4.84 4.86 4.88 4.9 4.92 4.94 4.96 4.98 5 5.2 5.4 5.6 5.8 6 6.2 6.4 6.6 6.8 7 7.2 7.4 7.6 7.8 8 8.2 8.4 8.6 8.8 9 9.2 9.4 9.6 9.8 10 11 12 13 14 15 20 1.00E+07

0.418586444 0.418262894 0.41794315 0.417627155 0.417314849 0.417006177 0.416701082 0.41639951 0.416101408 0.415806723 0.415515404 0.415227399 0.412519192 0.410093218 0.407911868 0.40594356 0.404161618 0.402543391 0.401069549 0.399723516 0.398491018 0.397359707 0.396318863 0.395359142 0.394472369 0.39365137 0.392889827 0.392182161 0.391523428 0.390909238 0.39033568 0.38979926 0.389296852 0.388825651 0.388383133 0.387967025 0.387575273 0.385922492 0.384661215 0.383677106 0.382894653 0.382262375 0.380387347 0.377964473

13.64498036 13.86197793 14.08198772 14.30504281 14.53117654 14.76042251 14.99281457 15.22838685 15.46717372 15.70920983 15.95453008 16.20316967 18.88005879 21.92958513 25.39147028 29.30832295 33.72574463 38.69243541 44.26030002 50.48455396 57.42382972 65.14028317 73.69969986 83.17160152 93.62935255 105.1502665 117.8157125 131.7112223 146.9265961 163.5560098 181.6981213 201.4561774 222.9381199 246.256693 271.5295492 298.8793565 328.4339049 514.1223892 777.2297061 1140.131886 1629.206319 2275.165169 9278.597922 2.77867E+32

26.4898 26.71333333 26.9378 27.1632 27.38953333 27.6168 27.845 28.07413333 28.3042 28.5352 28.76713333 29 31.38 33.85333333 36.42 39.08 41.83333333 44.68 47.62 50.65333333 53.78 57 60.31333333 63.72 67.22 70.81333333 74.5 78.28 82.15333333 86.12 90.18 94.33333333 98.58 102.92 107.3533333 111.88 116.5 141 167.8333333 197 228.5 262.3333333 466.5 1.16667E+14

5.381110175 5.418416281 5.455877314 5.493493286 5.531264206 5.569190083 5.607270929 5.645506752 5.683897561 5.722443366 5.761144176 5.8 6.197085799 6.609681451 7.037793367 7.481426873 7.94058642 8.415275754 8.905498047 9.411255994 9.932551903 10.46938776 11.02176526 11.5896859 12.17315097 12.77216159 13.38671875 14.01682332 14.66247606 15.32367766 16.00042872 16.69272977 17.40058129 18.1239837 18.8629374 19.61744273 20.3875 24.47107438 28.94347994 33.80473373 39.05484694 44.69382716 78.721875 1.94444E+13

4.922738829 4.930099857 4.937391083 4.944613306 4.95176732 4.958853906 4.965873837 4.972827873 4.979716769 4.986541268 4.993302103 5 5.063670412 5.121779859 5.174917492 5.223602484 5.268292683 5.309392265 5.347258486 5.382207578 5.414519906 5.444444444 5.472202674 5.497991968 5.521988528 5.544349939 5.565217391 5.584717608 5.602964531 5.62006079 5.636098981 5.651162791 5.665327979 5.678663239 5.691230959 5.703087886 5.714285714 5.761904762 5.798657718 5.827586207 5.850746269 5.869565217 5.925925926 6

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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378

Appendix D Tables for Normal Shock Waves

γ = 5/3 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24 1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42 1.44 1.46 1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68 1.7 1.72 1.74 1.76 1.78 1.8 1.82 1.84 1.86 1.88 1.9 1.92 1.94

1 0.980582543 0.96226442 0.944955378 0.928575067 0.913051678 0.898320807 0.884324504 0.871010458 0.858331311 0.846244074 0.834709616 0.823692236 0.813159283 0.803080831 0.793429393 0.784179673 0.775308347 0.766793872 0.758616312 0.750757194 0.743199367 0.735926889 0.72892492 0.722179623 0.715678085 0.709408236 0.703358781 0.69751914 0.691879391 0.686430215 0.681162859 0.676069086 0.671141138 0.666371706 0.661753894 0.657281189 0.65294744 0.648746827 0.644673845 0.640723276 0.636890176 0.633169857 0.629557868 0.626049982 0.62264218 0.619330643 0.616111734

1 1.000009565 1.000073271 1.00023703 1.00053907 1.001011152 1.001679587 1.002566103 1.003688558 1.00506156 1.006696976 1.008604372 1.010791377 1.013263999 1.016026884 1.019083545 1.022436547 1.026087672 1.030038052 1.034288288 1.038838546 1.043688642 1.048838109 1.054286265 1.060032252 1.066075091 1.072413707 1.07904697 1.085973709 1.093192746 1.100702902 1.108503019 1.116591973 1.124968676 1.133632094 1.142581246 1.151815213 1.161333141 1.171134245 1.181217807 1.191583183 1.202229801 1.213157161 1.224364837 1.235852473 1.247619786 1.259666563 1.271992661

1 1.0505 1.102 1.1545 1.208 1.2625 1.318 1.3745 1.432 1.4905 1.55 1.6105 1.672 1.7345 1.798 1.8625 1.928 1.9945 2.062 2.1305 2.2 2.2705 2.342 2.4145 2.488 2.5625 2.638 2.7145 2.792 2.8705 2.95 3.0305 3.112 3.1945 3.278 3.3625 3.448 3.5345 3.622 3.7105 3.8 3.8905 3.982 4.0745 4.168 4.2625 4.358 4.4545

1 1.019905854 1.03964571 1.059250667 1.078748971 1.098166322 1.117526148 1.136849838 1.156156956 1.175465419 1.194791667 1.2141508 1.233556712 1.253022203 1.272559082 1.292178254 1.311889807 1.33170308 1.35162673 1.371668793 1.391836735 1.412137498 1.432577546 1.453162906 1.473899196 1.494791667 1.515845222 1.53706445 1.558453649 1.580016844 1.601757813 1.623680098 1.645787032 1.668081743 1.690567177 1.713246107 1.736121147 1.759194758 1.782469267 1.805946866 1.82962963 1.853519518 1.877618384 1.901927983 1.926449977 1.951185942 1.97613737 2.001305678

1 1.02999703 1.05997648 1.089921428 1.119815668 1.149643705 1.179390748 1.209042702 1.238586156 1.268008378 1.297297297 1.326441494 1.355430183 1.384253204 1.412901 1.441364606 1.469635628 1.497706231 1.525569119 1.553217519 1.580645161 1.607846264 1.634815516 1.661548055 1.688039457 1.714285714 1.740283218 1.766028744 1.791519435 1.816752784 1.841726619 1.866439087 1.890888639 1.915074015 1.938994229 1.962648557 1.98603652 2.009157874 2.032012595 2.054600869 2.076923077 2.098979786 2.120771736 2.142299833 2.163565132 2.184568835 2.205312276 2.225796913

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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Appendix D Tables for Normal Shock Waves

379

(continued) γ = 5/3 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

1.96 1.98 2 2.02 2.04 2.06 2.08 2.1 2.12 2.14 2.16 2.18 2.2 2.22 2.24 2.26 2.28 2.3 2.32 2.34 2.36 2.38 2.4 2.42 2.44 2.46 2.48 2.5 2.52 2.54 2.56 2.58 2.6 2.62 2.64 2.66 2.68 2.7 2.72 2.74 2.76 2.78 2.8 2.82 2.84 2.86 2.88

0.612981994 0.609938123 0.606976979 0.604095564 0.601291018 0.598560611 0.595901733 0.593311893 0.590788707 0.588329896 0.585933279 0.583596766 0.581318359 0.57909614 0.576928272 0.574812993 0.572748612 0.570733509 0.568766126 0.566844966 0.564968593 0.563135626 0.561344737 0.559594649 0.557884132 0.556212006 0.554577131 0.552978412 0.551414792 0.549885254 0.548388817 0.546924536 0.545491499 0.544088826 0.542715669 0.541371209 0.540054653 0.538765239 0.537502229 0.53626491 0.535052593 0.533864611 0.532700322 0.531559103 0.53044035 0.529343482 0.528267933

1.284598004 1.297482584 1.310646457 1.324089743 1.337812624 1.351815343 1.3660982 1.380661553 1.395505815 1.410631452 1.426038983 1.441728977 1.45770205 1.473958867 1.490500137 1.507326614 1.524439093 1.541838412 1.559525447 1.577501113 1.595766362 1.614322181 1.633169593 1.652309652 1.671743446 1.691472093 1.711496741 1.731818568 1.752438777 1.773358602 1.794579299 1.816102152 1.837928468 1.860059576 1.88249683 1.905241605 1.928295297 1.951659321 1.975335114 1.999324129 2.023627841 2.04824774 2.073185334 2.098442147 2.124019719 2.149919608 2.176143382

4.552 4.6505 4.75 4.8505 4.952 5.0545 5.158 5.2625 5.368 5.4745 5.582 5.6905 5.8 5.9105 6.022 6.1345 6.248 6.3625 6.478 6.5945 6.712 6.8305 6.95 7.0705 7.192 7.3145 7.438 7.5625 7.688 7.8145 7.942 8.0705 8.2 8.3305 8.462 8.5945 8.728 8.8625 8.998 9.1345 9.272 9.4105 9.55 9.6905 9.832 9.9745 10.118

2.026692212 2.052298248 2.078125 2.104173623 2.130445213 2.156940817 2.183661428 2.210607993 2.237781417 2.26518256 2.292812243 2.32067125 2.348760331 2.377080198 2.405631537 2.434414999 2.46343121 2.492680766 2.522164239 2.551882177 2.581835105 2.612023524 2.642447917 2.673108744 2.70400645 2.735141458 2.766514178 2.798125 2.829974301 2.862062442 2.894389771 2.926956621 2.959763314 2.992810158 3.026097452 3.059625481 3.09339452 3.127404835 3.161656683 3.196150308 3.230885948 3.265863833 3.301084184 3.336547212 3.372253124 3.408202119 3.444394387

2.246024322 2.265996185 2.285714286 2.305180498 2.324396783 2.343365178 2.362087792 2.380566802 2.398804441 2.416802998 2.434564809 2.452092255 2.469387755 2.486453761 2.503292756 2.519907248 2.536299766 2.552472859 2.56842909 2.584171032 2.59970127 2.61502239 2.630136986 2.645047649 2.659756969 2.674267533 2.68858192 2.702702703 2.716632444 2.730373693 2.74392899 2.757300857 2.770491803 2.783504319 2.796340876 2.80900393 2.821495915 2.833819242 2.845976304 2.85796947 2.869801085 2.881473472 2.89298893 2.904349732 2.915558126 2.926616337 2.937526562

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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380

Appendix D Tables for Normal Shock Waves

(continued) γ = 5/3 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

2.9 2.92 2.94 2.96 2.98 3 3.02 3.04 3.06 3.08 3.1 3.12 3.14 3.16 3.18 3.2 3.22 3.24 3.26 3.28 3.3 3.32 3.34 3.36 3.38 3.4 3.42 3.44 3.46 3.48 3.5 3.52 3.54 3.56 3.58 3.6 3.62 3.64 3.66 3.68 3.7 3.72 3.74 3.76 3.78 3.8 3.82

0.527213159 0.52617863 0.525163835 0.524168277 0.523191477 0.522232968 0.5212923 0.520369034 0.519462748 0.51857303 0.517699481 0.516841713 0.515999352 0.515172033 0.514359402 0.513561116 0.512776841 0.512006254 0.511249039 0.510504892 0.509773515 0.509054619 0.508347925 0.507653159 0.506970056 0.506298359 0.505637816 0.504988184 0.504349225 0.503720708 0.503102408 0.502494108 0.501895593 0.501306656 0.500727096 0.500156715 0.499595323 0.499042732 0.498498762 0.497963234 0.497435976 0.49691682 0.496405601 0.495902161 0.495406342 0.494917993 0.494436966

2.202692629 2.229568946 2.256773947 2.284309258 2.312176517 2.340377375 2.368913494 2.397786549 2.426998226 2.456550219 2.486444236 2.516681992 2.547265215 2.578195641 2.609475013 2.641105086 2.673087623 2.705424394 2.738117178 2.771167763 2.804577942 2.838349517 2.872484299 2.906984101 2.941850748 2.977086069 3.012691899 3.04867008 3.085022461 3.121750894 3.15885724 3.196343363 3.234211133 3.272462426 3.311099123 3.350123108 3.389536272 3.42934051 3.469537721 3.510129809 3.551118682 3.592506251 3.634294434 3.67648515 3.719080322 3.76208188 3.805491754

10.2625 10.408 10.5545 10.702 10.8505 11 11.1505 11.302 11.4545 11.608 11.7625 11.918 12.0745 12.232 12.3905 12.55 12.7105 12.872 13.0345 13.198 13.3625 13.528 13.6945 13.862 14.0305 14.2 14.3705 14.542 14.7145 14.888 15.0625 15.238 15.4145 15.592 15.7705 15.95 16.1305 16.312 16.4945 16.678 16.8625 17.048 17.2345 17.422 17.6105 17.8 17.9905

3.480830113 3.517509476 3.55443265 3.591599799 3.629011086 3.666666667 3.704566691 3.742711305 3.78110065 3.819734863 3.858614074 3.897738412 3.937108001 3.976722961 4.016583407 4.056689453 4.097041207 4.137638775 4.178482258 4.219571758 4.260907369 4.302489186 4.344317298 4.386391794 4.42871276 4.471280277 4.514094426 4.557155287 4.600462933 4.64401744 4.687818878 4.731867317 4.776162824 4.820705466 4.865495307 4.910532407 4.955816829 5.001348629 5.047127867 5.093154596 5.139428871 5.185950746 5.23272027 5.279737494 5.327002467 5.374515235 5.422275846

2.948290973 2.958911716 2.96939091 2.979730649 2.989932999 3 3.009933666 3.019735982 3.02940891 3.038954382 3.048374306 3.057670562 3.066845003 3.075899458 3.084835728 3.093655589 3.102360791 3.110953058 3.11943409 3.127805559 3.136069114 3.144226381 3.152278957 3.160228418 3.168076315 3.175824176 3.183473504 3.191025779 3.19848246 3.205844981 3.213114754 3.22029317 3.227381596 3.23438138 3.241293847 3.248120301 3.254862025 3.261520284 3.268096319 3.274591353 3.281006591 3.287343216 3.293602392 3.299785267 3.305892969 3.311926606 3.31788727

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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Appendix D Tables for Normal Shock Waves

381

(continued) γ = 5/3 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

3.84 3.86 3.88 3.9 3.92 3.94 3.96 3.98 4 4.02 4.04 4.06 4.08 4.1 4.12 4.14 4.16 4.18 4.2 4.22 4.24 4.26 4.28 4.3 4.32 4.34 4.36 4.38 4.4 4.42 4.44 4.46 4.48 4.5 4.52 4.54 4.56 4.58 4.6 4.62 4.64 4.66 4.68 4.7 4.72 4.74 4.76

0.493963116 0.493496302 0.493036385 0.492583233 0.492136713 0.491696698 0.491263062 0.490835685 0.490414446 0.48999923 0.489589923 0.489186415 0.488788596 0.488396361 0.488009607 0.487628233 0.48725214 0.486881232 0.486515414 0.486154594 0.485798683 0.485447591 0.485101234 0.484759527 0.484422387 0.484089734 0.483761489 0.483437576 0.483117918 0.482802442 0.482491076 0.482183749 0.481880392 0.481580938 0.481285319 0.480993472 0.480705332 0.480420838 0.480139929 0.479862544 0.479588627 0.479318119 0.479050964 0.478787108 0.478526496 0.478269077 0.478014799

3.849311878 3.893544192 3.938190636 3.983253156 4.028733699 4.074634217 4.120956664 4.167702997 4.214875175 4.262475162 4.310504922 4.358966425 4.40786164 4.457192542 4.506961104 4.557169306 4.607819128 4.658912553 4.710451564 4.762438151 4.814874301 4.867762006 4.921103258 4.974900055 5.029154392 5.083868268 5.139043685 5.194682645 5.250787152 5.307359213 5.364400835 5.421914029 5.479900804 5.538363173 5.597303151 5.656722753 5.716623997 5.7770089 5.837879482 5.899237765 5.961085772 6.023425526 6.086259051 6.149588376 6.213415527 6.277742532 6.342571423

18.182 18.3745 18.568 18.7625 18.958 19.1545 19.352 19.5505 19.75 19.9505 20.152 20.3545 20.558 20.7625 20.968 21.1745 21.382 21.5905 21.8 22.0105 22.222 22.4345 22.648 22.8625 23.078 23.2945 23.512 23.7305 23.95 24.1705 24.392 24.6145 24.838 25.0625 25.288 25.5145 25.742 25.9705 26.2 26.4305 26.662 26.8945 27.128 27.3625 27.598 27.8345 28.072

5.470284342 5.51854077 5.56704517 5.615797584 5.664798053 5.714046616 5.763543312 5.813288178 5.86328125 5.913522564 5.964012156 6.014750058 6.065736303 6.116970925 6.168453954 6.220185421 6.272165357 6.32439379 6.376870748 6.429596261 6.482570354 6.535793055 6.58926439 6.642984383 6.696953061 6.751170446 6.805636563 6.860351434 6.915315083 6.970527531 7.0259888 7.081698911 7.137657884 7.193865741 7.2503225 7.307028181 7.363982802 7.421186383 7.478638941 7.536340494 7.59429106 7.652490654 7.710939294 7.769636996 7.828583776 7.887779649 7.947224631

3.323776035 3.329593957 3.335342077 3.341021417 3.346632982 3.352177762 3.357656732 3.363070848 3.368421053 3.373708273 3.378933422 3.384097395 3.389201075 3.394245331 3.399231016 3.40415897 3.409030021 3.413844982 3.418604651 3.423309817 3.427961254 3.432559723 3.437105974 3.441600745 3.44604476 3.450438733 3.454783367 3.459079353 3.46332737 3.467528088 3.471682164 3.475790246 3.479852972 3.483870968 3.487844851 3.491775229 3.495662699 3.499507849 3.503311258 3.507073495 3.510795121 3.514476687 3.518118736 3.521721802 3.525286411 3.52881308 3.53230232

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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382

Appendix D Tables for Normal Shock Waves

(continued) γ = 5/3 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

4.78 4.8 4.82 4.84 4.86 4.88 4.9 4.92 4.94 4.96 4.98 5 5.2 5.4 5.6 5.8 6 6.2 6.4 6.6 6.8 7 7.2 7.4 7.6 7.8 8 8.2 8.4 8.6 8.8 9 9.2 9.4 9.6 9.8 10 11 12 13 14 15 20 1.00E+07

0.47776361 0.477515461 0.477270305 0.477028092 0.476788777 0.476552313 0.476318656 0.476087762 0.475859588 0.475634091 0.47541123 0.475190963 0.473122614 0.471274281 0.469615963 0.468122586 0.466773066 0.46554957 0.464436937 0.463422213 0.462494276 0.461643536 0.460861686 0.460141506 0.459476694 0.458861733 0.458291772 0.457762536 0.457270245 0.456811547 0.45638346 0.455983324 0.455608764 0.45525765 0.454928069 0.454618297 0.454326783 0.453098238 0.452162168 0.451432685 0.450853237 0.450385363 0.448999778 0.447213595

6.40790423 6.473742985 6.540089721 6.606946474 6.674315278 6.742198169 6.810597186 6.879514367 6.94895175 7.018911377 7.089395289 7.160405527 7.899905771 8.69429113 9.545616174 10.45594136 11.42733204 12.46185765 13.56159104 14.72860797 15.96498669 17.27280751 18.65415258 20.1111056 21.64575162 23.26017686 24.95646857 26.73671489 28.60300477 30.55742782 32.60207431 34.73903501 36.97040121 39.29826462 41.72471734 44.25185182 46.88176082 61.64618847 79.29402747 100.0870895 124.2872518 152.1564348 355.7065504 4.36732E+19

28.3105 28.55 28.7905 29.032 29.2745 29.518 29.7625 30.008 30.2545 30.502 30.7505 31 33.55 36.2 38.95 41.8 44.75 47.8 50.95 54.2 57.55 61 64.55 68.2 71.95 75.8 79.75 83.8 87.95 92.2 96.55 101 105.55 110.2 114.95 119.8 124.75 151 179.75 211 244.75 281 499.75 1.25E+14

8.006918736 8.066861979 8.127054374 8.187495936 8.248186678 8.309126612 8.370315754 8.431754115 8.493441707 8.555378544 8.617564638 8.68 9.318065828 9.981069959 10.66902105 11.38192628 12.11979167 12.88262227 13.67042236 14.48319559 15.32094507 16.18367347 17.0713831 17.98407597 18.92175381 19.88441815 20.87207031 21.88471148 22.92234269 23.98496485 25.07257877 26.18518519 27.32278474 28.485378 29.67296549 30.88554769 32.123125 38.68595041 45.87369792 53.68639053 62.12404337 71.18666667 125.8745313 3.125E+13

3.535754631 3.539170507 3.542550434 3.54589489 3.549204346 3.552479265 3.555720104 3.558927311 3.56210133 3.565242595 3.568351535 3.571428571 3.600532623 3.626865672 3.650756694 3.672489083 3.692307692 3.71042471 3.727024568 3.742268041 3.756295695 3.769230769 3.781181619 3.792243767 3.802501646 3.812030075 3.820895522 3.829157175 3.836867863 3.844074844 3.850820487 3.857142857 3.863076221 3.868651489 3.873896595 3.878836834 3.883495146 3.903225806 3.918367347 3.930232558 3.939698492 3.947368421 3.970223325 4

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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Appendix D Tables for Normal Shock Waves

383

γ = 1.3 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24 1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42 1.44 1.46 1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68 1.7 1.72 1.74 1.76 1.78 1.8 1.82 1.84 1.86 1.88 1.9 1.92 1.94

1 0.980491952 0.961920674 0.944220566 0.927332174 0.911201473 0.895779238 0.88102052 0.866884173 0.853332454 0.840330669 0.827846867 0.815851566 0.804317513 0.793219477 0.782534057 0.772239517 0.762315637 0.752743583 0.743505785 0.734585834 0.72596838 0.717639055 0.709584388 0.701791736 0.694249222 0.686945676 0.679870583 0.673014035 0.666366684 0.659919708 0.653664771 0.647593988 0.641699898 0.635975434 0.630413896 0.625008929 0.619754499 0.614644877 0.609674613 0.604838524 0.600131679 0.595549378 0.591087143 0.586740706 0.582505991 0.578379109 0.574356346

1 1.0000101 1.000077901 1.000253686 1.000580688 1.001096091 1.001831897 1.002815672 1.004071191 1.005618988 1.007476845 1.009660195 1.012182486 1.015055485 1.018289545 1.021893834 1.025876535 1.030245014 1.035005971 1.040165567 1.045729536 1.05170328 1.058091951 1.064900527 1.072133871 1.079796785 1.08789406 1.096430515 1.105411033 1.114840592 1.12472429 1.135067372 1.145875248 1.157153514 1.168907961 1.181144598 1.193869657 1.207089606 1.220811162 1.235041294 1.249787234 1.265056482 1.280856816 1.297196294 1.314083259 1.331526347 1.349534491 1.368116923

1 1.045669565 1.092243478 1.139721739 1.188104348 1.237391304 1.287582609 1.338678261 1.390678261 1.443582609 1.497391304 1.552104348 1.607721739 1.664243478 1.721669565 1.78 1.839234783 1.899373913 1.960417391 2.022365217 2.085217391 2.148973913 2.213634783 2.2792 2.345669565 2.413043478 2.481321739 2.550504348 2.620591304 2.691582609 2.763478261 2.836278261 2.909982609 2.984591304 3.060104348 3.136521739 3.213843478 3.292069565 3.3712 3.451234783 3.532173913 3.614017391 3.696765217 3.780417391 3.864973913 3.950434783 4.0368 4.124069565

1 1.010361197 1.020588709 1.030701349 1.040716202 1.050648815 1.06051336 1.070322776 1.080088896 1.08982256 1.099533711 1.109231486 1.118924287 1.128619858 1.138325343 1.148047337 1.157791944 1.167564812 1.177371179 1.187215905 1.197103507 1.207038184 1.217023847 1.22706414 1.237162463 1.247321991 1.257545692 1.26783634 1.278196536 1.288628715 1.299135161 1.309718018 1.320379301 1.331120903 1.341944606 1.352852088 1.363844928 1.374924617 1.386092562 1.397350089 1.408698453 1.420138839 1.431672368 1.443300103 1.455023049 1.466842158 1.478758333 1.490772433

1 1.034946283 1.070209251 1.105773016 1.141621842 1.177740161 1.214112577 1.250723886 1.287559076 1.324603345 1.361842105 1.399260991 1.436845869 1.474582842 1.512458258 1.550458716 1.588571066 1.626782422 1.66508016 1.703451924 1.741885626 1.780369455 1.818891872 1.857441617 1.896007708 1.934579439 1.973146388 2.011698408 2.050225634 2.088718479 2.12716763 2.165564054 2.203898991 2.242163952 2.280350719 2.318451343 2.356458137 2.394363679 2.432160804 2.469842604 2.507402423 2.544833852 2.58213073 2.619287134 2.656297381 2.693156017 2.72985782 2.766397791

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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384

Appendix D Tables for Normal Shock Waves

(continued) γ = 1.3 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

1.96 1.98 2 2.02 2.04 2.06 2.08 2.1 2.12 2.14 2.16 2.18 2.2 2.22 2.24 2.26 2.28 2.3 2.32 2.34 2.36 2.38 2.4 2.42 2.44 2.46 2.48 2.5 2.52 2.54 2.56 2.58 2.6 2.62 2.64 2.66 2.68 2.7 2.72 2.74 2.76 2.78 2.8 2.82 2.84 2.86 2.88

0.570434152 0.566609131 0.562878036 0.559237759 0.555685323 0.552217877 0.548832689 0.545527137 0.542298707 0.539144988 0.536063662 0.533052506 0.530109381 0.527232233 0.524419086 0.521668039 0.518977265 0.516345001 0.513769553 0.511249288 0.508782632 0.506368069 0.504004136 0.501689422 0.499422567 0.497202257 0.495027225 0.492896245 0.490808136 0.488761754 0.486755996 0.484789794 0.482862116 0.480971963 0.47911837 0.477300401 0.475517153 0.47376775 0.472051343 0.47036711 0.468714257 0.467092012 0.465499627 0.463936377 0.462401561 0.460894497 0.459414523

1.38728318 1.407043107 1.427406862 1.448384919 1.469988073 1.492227441 1.515114469 1.538660932 1.562878941 1.587780944 1.613379732 1.639688443 1.666720562 1.694489929 1.723010741 1.752297558 1.782365303 1.813229273 1.844905136 1.877408939 1.910757114 1.944966479 1.980054247 2.016038024 2.052935824 2.090766063 2.129547573 2.169299601 2.210041819 2.251794326 2.294577655 2.338412779 2.383321116 2.429324535 2.47644536 2.524706382 2.574130857 2.624742518 2.67656558 2.729624746 2.783945213 2.839552679 2.896473349 2.954733947 3.014361713 3.07538442 3.137830374

4.212243478 4.301321739 4.391304348 4.482191304 4.573982609 4.666678261 4.760278261 4.854782609 4.950191304 5.046504348 5.143721739 5.241843478 5.340869565 5.4408 5.541634783 5.643373913 5.746017391 5.849565217 5.954017391 6.059373913 6.165634783 6.2728 6.380869565 6.489843478 6.599721739 6.710504348 6.822191304 6.934782609 7.048278261 7.162678261 7.277982609 7.394191304 7.511304348 7.629321739 7.748243478 7.868069565 7.9888 8.110434783 8.232973913 8.356417391 8.480765217 8.606017391 8.732173913 8.859234783 8.9872 9.116069565 9.245843478

1.50288527 1.515097617 1.527410208 1.539823742 1.552338881 1.56495626 1.577676478 1.590500109 1.6034277 1.616459771 1.62959682 1.64283932 1.656187724 1.669642464 1.683203954 1.696872589 1.710648747 1.724532788 1.73852506 1.752625892 1.766835602 1.781154495 1.795582861 1.81012098 1.824769119 1.839527536 1.854396478 1.869376181 1.884466874 1.899668774 1.914982092 1.93040703 1.945943781 1.961592534 1.977353468 1.993226756 2.009212564 2.025311054 2.041522379 2.057846689 2.074284128 2.090834833 2.107498939 2.124276574 2.141167864 2.158172926 2.175291878

2.802771152 2.838973339 2.875 2.910846991 2.946510368 2.981986386 3.017271492 3.052362323 3.087255698 3.121948617 3.156438253 3.19072195 3.224797219 3.25866173 3.29231331 3.325749941 3.358969749 3.391971006 3.424752125 3.45731165 3.489648259 3.521760756 3.553648069 3.585309243 3.616743439 3.64794993 3.678928096 3.709677419 3.740197484 3.770487971 3.800548653 3.830379392 3.859980139 3.889350926 3.918491865 3.947403146 3.976085031 4.004537855 4.03276202 4.060757993 4.088526304 4.116067542 4.143382353 4.170471439 4.197335554 4.2239755 4.250392129

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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Appendix D Tables for Normal Shock Waves

385

(continued) γ = 1.3 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

2.9 2.92 2.94 2.96 2.98 3 3.02 3.04 3.06 3.08 3.1 3.12 3.14 3.16 3.18 3.2 3.22 3.24 3.26 3.28 3.3 3.32 3.34 3.36 3.38 3.4 3.42 3.44 3.46 3.48 3.5 3.52 3.54 3.56 3.58 3.6 3.62 3.64 3.66 3.68 3.7 3.72 3.74 3.76 3.78 3.8 3.82

0.457960998 0.456533301 0.455130826 0.453752987 0.452399215 0.451068956 0.449761675 0.448476847 0.447213967 0.445972542 0.444752092 0.443552152 0.442372268 0.441212001 0.440070922 0.438948613 0.43784467 0.436758698 0.435690313 0.43463914 0.433604816 0.432586986 0.431585305 0.430599437 0.429629054 0.428673837 0.427733475 0.426807664 0.42589611 0.424998525 0.424114627 0.423244144 0.422386807 0.421542356 0.420710538 0.419891105 0.419083814 0.418288431 0.417504724 0.416732468 0.415971445 0.415221441 0.414482245 0.413753654 0.413035469 0.412327494 0.411629539

3.201728426 3.267107977 3.333998985 3.402431973 3.472438038 3.544048857 3.617296696 3.692214417 3.768835486 3.847193981 3.927324601 4.009262675 4.093044168 4.17870569 4.266284508 4.355818549 4.447346413 4.540907381 4.636541424 4.734289209 4.834192114 4.936292231 5.040632382 5.147256121 5.25620775 5.367532324 5.481275665 5.597484367 5.71620581 5.837488168 5.96138042 6.08793236 6.217194605 6.349218609 6.484056674 6.621761954 6.762388475 6.905991139 7.052625735 7.202348957 7.355218405 7.511292604 7.670631012 7.833294032 7.999343025 8.168840316 8.341849215

9.376521739 9.508104348 9.640591304 9.773982609 9.908278261 10.04347826 10.17958261 10.3165913 10.45450435 10.59332174 10.73304348 10.87366957 11.0152 11.15763478 11.30097391 11.44521739 11.59036522 11.73641739 11.88337391 12.03123478 12.18 12.32966957 12.48024348 12.63172174 12.78410435 12.9373913 13.09158261 13.24667826 13.40267826 13.55958261 13.7173913 13.87610435 14.03572174 14.19624348 14.35766957 14.52 14.68323478 14.84737391 15.01241739 15.17836522 15.34521739 15.51297391 15.68163478 15.8512 16.02166957 16.19304348 16.36532174

2.192524832 2.209871895 2.227333172 2.244908763 2.262598767 2.280403277 2.298322384 2.316356177 2.334504741 2.352768159 2.37114651 2.389639871 2.408248318 2.426971924 2.445810757 2.464764887 2.483834378 2.503019297 2.522319703 2.541735657 2.561267218 2.580914441 2.600677383 2.620556095 2.64055063 2.660661037 2.680887366 2.701229663 2.721687975 2.742262345 2.762952818 2.783759435 2.804682238 2.825721265 2.846876556 2.868148148 2.889536078 2.911040381 2.932661091 2.954398243 2.976251869 2.998222001 3.02030867 3.042511906 3.064831738 3.087268195 3.109821305

4.276586337 4.302559062 4.328311286 4.353844027 4.379158341 4.404255319 4.429136086 4.453801797 4.478253637 4.502492819 4.526520582 4.550338189 4.573946929 4.597348109 4.620543058 4.643533123 4.66631967 4.68890408 4.711287749 4.733472086 4.755458515 4.77724847 4.798843394 4.820244743 4.841453977 4.862472568 4.88330199 4.903943727 4.924399265 4.944670094 4.964757709 4.984663607 5.004389285 5.023936244 5.043305982 5.0625 5.081519797 5.100366869 5.119042714 5.137548823 5.155886687 5.174057794 5.192063625 5.20990566 5.227585373 5.245104232 5.262463702

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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386

Appendix D Tables for Normal Shock Waves

(continued) γ = 1.3 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

3.84 3.86 3.88 3.9 3.92 3.94 3.96 3.98 4 4.02 4.04 4.06 4.08 4.1 4.12 4.14 4.16 4.18 4.2 4.22 4.24 4.26 4.28 4.3 4.32 4.34 4.36 4.38 4.4 4.42 4.44 4.46 4.48 4.5 4.52 4.54 4.56 4.58 4.6 4.62 4.64 4.66 4.68 4.7 4.72 4.74 4.76

0.410941418 0.41026295 0.409593957 0.408934265 0.408283706 0.407642113 0.407009324 0.406385181 0.405769529 0.405162216 0.404563095 0.403972021 0.403388851 0.402813448 0.402245675 0.401685401 0.401132495 0.400586829 0.400048281 0.399516727 0.39899205 0.398474132 0.397962858 0.397458119 0.396959802 0.396467803 0.395982015 0.395502336 0.395028664 0.394560902 0.394098953 0.393642721 0.393192114 0.392747042 0.392307414 0.391873143 0.391444145 0.391020334 0.390601629 0.390187948 0.389779213 0.389375347 0.388976272 0.388581915 0.388192202 0.387807061 0.387426422

8.518434019 8.698660032 8.882593571 9.070301982 9.261853649 9.457318011 9.656765568 9.8602679 10.06789767 10.27972866 10.49583575 10.71629495 10.94118342 11.17057946 11.40456256 11.64321336 11.88661373 12.13484672 12.3879966 12.64614891 12.90939039 13.17780908 13.45149428 13.73053658 14.01502789 14.30506144 14.60073176 14.90213477 15.20936774 15.52252931 15.84171952 16.16703982 16.49859309 16.83648362 17.1808172 17.53170104 17.88924388 18.25355594 18.62474894 19.00293617 19.38823245 19.78075415 20.18061925 20.58794731 21.0028595 21.42547865 21.85592919

16.53850435 16.7125913 16.88758261 17.06347826 17.24027826 17.41798261 17.5965913 17.77610435 17.95652174 18.13784348 18.32006957 18.5032 18.68723478 18.87217391 19.05801739 19.24476522 19.43241739 19.62097391 19.81043478 20.0008 20.19206957 20.38424348 20.57732174 20.77130435 20.9661913 21.16198261 21.35867826 21.55627826 21.75478261 21.9541913 22.15450435 22.35572174 22.55784348 22.76086957 22.9648 23.16963478 23.37537391 23.58201739 23.78956522 23.99801739 24.20737391 24.41763478 24.6288 24.84086957 25.05384348 25.26772174 25.48250435

3.132491096 3.155277592 3.178180821 3.201200807 3.224337575 3.247591148 3.27096155 3.294448803 3.31805293 3.341773952 3.36561189 3.389566765 3.413638596 3.437827403 3.462133205 3.486556021 3.511095868 3.535752765 3.560526729 3.585417776 3.610425923 3.635551187 3.660793582 3.686153124 3.711629829 3.73722371 3.762934783 3.78876306 3.814708557 3.840771285 3.866951259 3.89324849 3.919662992 3.946194777 3.972843856 3.999610242 4.026493945 4.053494977 4.080613348 4.10784907 4.135202153 4.162672607 4.190260443 4.217965669 4.245788296 4.273728332 4.301785788

5.279665239 5.296710294 5.313600314 5.330336736 5.346920991 5.363354504 5.379638689 5.395774957 5.411764706 5.427609329 5.44331021 5.458868724 5.474286237 5.489564106 5.504703679 5.519706296 5.534573285 5.549305966 5.56390565 5.578373637 5.592711219 5.606919675 5.621000277 5.634954286 5.648782953 5.662487517 5.67606921 5.689529252 5.702868852 5.716089211 5.729191517 5.742176949 5.755046677 5.767801858 5.78044364 5.792973161 5.805391548 5.817699918 5.829899377 5.841991023 5.85397594 5.865855205 5.877629884 5.889301032 5.900869693 5.912336905 5.92370369

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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Appendix D Tables for Normal Shock Waves

387

(continued) γ = 1.3 M1

M2

Pt1 =Pt2 ¼ A2 =A1

P2 =P1

T2 =T1

ρ2 =ρ1 ¼ V1n =V2n

Tt2 =Tt1

4.78 4.8 4.82 4.84 4.86 4.88 4.9 4.92 4.94 4.96 4.98 5 5.2 5.4 5.6 5.8 6 6.2 6.4 6.6 6.8 7 7.2 7.4 7.6 7.8 8 8.2 8.4 8.6 8.8 9 9.2 9.4 9.6 9.8 10 11 12 13 14 15 20 1.00E+07

0.387050216 0.386678376 0.386310834 0.385947526 0.385588387 0.385233354 0.384882367 0.384535364 0.384192286 0.383853075 0.383517674 0.383186026 0.380064322 0.377263109 0.374740378 0.372460733 0.370394188 0.36851521 0.366801955 0.365235643 0.363800067 0.362481178 0.361266753 0.360146121 0.359109928 0.358149953 0.357258947 0.356430497 0.355658913 0.354939137 0.354266657 0.35363744 0.35304787 0.352494702 0.351975013 0.351486167 0.351025781 0.349081792 0.347596475 0.346436457 0.345513445 0.344767131 0.342551516 0.33968311

22.29433726 22.74083065 23.19553887 23.65859312 24.13012637 24.6102733 25.09917039 25.59695588 26.10376984 26.61975413 27.14505247 27.67981043 33.5810597 40.60290575 48.92453205 58.74800005 70.30039548 83.8361021 99.63920616 118.0260343 139.347828 163.9935576 192.3928785 225.019233 262.3930998 305.0853948 353.7210255 408.982602 471.6143073 542.4259299 622.2970602 712.1814559 813.1115759 926.2032879 1052.660751 1193.781477 1350.961571 2434.216918 4195.090842 6955.072479 11146.48781 17338.31191 111792.5101 1.02531E+43

25.6981913 25.91478261 26.13227826 26.35067826 26.56998261 26.7901913 27.01130435 27.23332174 27.45624348 27.68006957 27.9048 28.13043478 30.43652174 32.83304348 35.32 37.8973913 40.56521739 43.32347826 46.17217391 49.11130435 52.14086957 55.26086957 58.47130435 61.77217391 65.16347826 68.64521739 72.2173913 75.88 79.63304348 83.47652174 87.41043478 91.43478261 95.54956522 99.75478261 104.0504348 108.4365217 112.9130435 136.6521739 162.6521739 190.9130435 221.4347826 254.2173913 452.0434783 1.13043E+14

4.329960672 4.358252993 4.38666276 4.41518998 4.443834663 4.472596817 4.501476448 4.530473566 4.559588178 4.588820292 4.618169913 4.647637051 4.94877328 5.261668028 5.586326531 5.922753136 6.270951481 6.630924624 7.002675154 7.386205273 7.781516866 8.18861155 8.607490723 9.038155595 9.480607219 9.934846516 10.40087429 10.87869126 11.36829803 11.86969518 12.38288319 12.90786249 13.44463349 13.99319652 14.55355191 15.12569992 15.70964083 18.806246 22.19770006 25.88401696 29.86520582 34.14127284 59.94489603 1.47448E+13

5.934971066 5.946140036 5.957211596 5.968186732 5.97906642 5.989851624 6.000543301 6.011142398 6.021649852 6.032066589 6.042393529 6.052631579 6.150316456 6.240044659 6.322580645 6.398610652 6.46875 6.533550103 6.593505039 6.649057606 6.700604839 6.748502994 6.793072015 6.834599522 6.873344371 6.909539799 6.943396226 6.975103734 7.004834254 7.032743509 7.058972733 7.08365019 7.106892523 7.128805949 7.149487318 7.169025055 7.1875 7.266318538 7.327433628 7.375711575 7.414473684 7.446043165 7.540983607 7.666666667

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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APPENDIX E

Oblique-Shock Chart Note: The oblique-shock chart for γ = 1.4 is taken from NACA Report 1135 (Equations, Tables, and Charts for Compressible Flow, by Ames Research Staff, 1953).

90

8

M1 = 10 8 6 4.5 5 4.0 3.8 3.6 3.4 3.2 3.0

1.05

80

1.10 1.15 1.20 1.25

70

1.30

1.35

1.40

2.8 1.45 1.5

1.6

1.7

1.8

1.9

2.0

2.4

2.2

2.6

2.8

60

3.2 3.4 3.6 3.8 4.0 4.5 5 6 8 10 20

50

40

8

Shock-wave angle, θ, degrees

3.0

30

20

Weak shock wave Strong shock wave Sonic limit (M2 = 1)

10

0

0

4

8

12

16

20

24

28

32

Deflection angle, δ, degrees

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389

Appendix E Oblique-Shock Chart 90

80

70 2.6

3.6 2.8 3.0 3.2 3.4

8 6

20 10

8

M1=2.2 2.4

4.0 4.5 5 3.8

Shock-wave angle, θ, degrees

60

50

40

30 Shock wave M1 Streamline

M2 θ

δ

20

10

0 26

30

34

38 42 46 Deflection angle, δ, degrees

50

54

58

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APPENDIX F

Table for Fanno Line Flow

Note – The numerical values of the table can be directly obtained from the following equations. They are tabulated for γ ¼ 1:4 for convenience. (The subscripts t for thermodynamic properties denote stagnation values; the superscript * denotes critical condition for M ¼ 1.) !

γþ1 2 fL γþ1 M2  1 2 M ¼ (6.21a) ln  γ1 2γ γM2 Dh 1þ M2 2

#1=2 " 1 ðγ þ 1Þ=2 M 1 þ γ1 M2

(6.26)



 γþ1 1 2 γ  1 2 2ðγ1Þ 1þ M M γþ1 2

(6.27)

p ¼ p

pt ¼ pt

2

T ðγ þ 1Þ=2 ¼  2 T 1 þ γ1 2 M ρ V ¼ ¼ ρ V



 1 2 ðγ  1Þ 2 1=2 1þ M M γþ1 2

(6.28)

(6.29)

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391

Appendix F Table for Fanno Line Flow

γ = 1.4 Mach Number

fL*/Dh

p/p*

pt/pt*

T/T*

ρ/ρ*

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9 0.92 0.94

1778.44988 440.352214 193.031082 106.718216 66.92156 45.407962 32.511306 24.19783 18.5426545 14.5332665 11.5960544 9.38648053 7.68756658 6.35721449 5.29925311 4.44674347 3.75195255 3.18011752 2.70544783 2.30849265 1.97436603 1.69152484 1.45091135 1.24534129 1.06906031 0.91741799 0.78662509 0.67357071 0.57568302 0.49082205 0.41719657 0.35329885 0.29785323 0.24977519 0.20813851 0.17214885 0.14112224 0.11446756 0.09167216 0.07228997 0.05593167 0.04225644 0.03096515 0.02179454 0.01451239 0.00891334 0.00481545

54.770065 27.3817471 18.2508495 13.6843088 10.9435131 9.11559228 7.80931667 6.82907187 6.0661835 5.45544726 4.95536975 4.53828896 4.18505446 3.88198758 3.61905747 3.38874115 3.1852859 3.00421752 2.84200386 2.69581933 2.56337663 2.44280443 2.33255695 2.23134609 2.13808994 2.05187308 1.97191578 1.89754974 1.8281989 1.76336404 1.70261038 1.64555755 1.59187127 1.54125665 1.4934525 1.44822667 1.40537213 1.36470364 1.32605498 1.28927656 1.25423336 1.22080325 1.18887545 1.15834925 1.12913287 1.10114253 1.07430155

28.9421302 14.4814859 9.66591007 7.26160964 5.82182875 4.86431765 4.1823998 3.67273863 3.27792645 2.96352 2.7076021 2.49556245 2.31728731 2.16555358 2.03506526 1.92185128 1.82287575 1.73577828 1.65869617 1.59014 1.5289048 1.47400537 1.42462855 1.38009735 1.33984375 1.30338776 1.2703211 1.24029444 1.21300721 1.18819951 1.16564554 1.14514831 1.12653524 1.10965464 1.09437268 1.08057094 1.06814435 1.05699943 1.04705281 1.03823 1.03046428 1.02369581 1.01787085 1.01294105 1.00886287 1.00559707 1.00310828

1.2 1.19990401 1.19961612 1.19913662 1.19846596 1.19760479 1.19655392 1.19531437 1.1938873 1.19227406 1.19047619 1.18849536 1.18633344 1.18399242 1.18147448 1.17878193 1.17591722 1.17288295 1.16968185 1.16631677 1.1627907 1.15910672 1.15526802 1.15127792 1.1471398 1.14285714 1.13843352 1.13387255 1.12917796 1.1243535 1.11940299 1.11433029 1.10913931 1.10383398 1.09841828 1.09289617 1.08727167 1.08154878 1.0757315 1.06982384 1.06382979 1.05775333 1.05159843 1.04536902 1.03906899 1.03270224 1.02627258 1.01978381

45.6453722 22.8254244 15.2199917 11.4181872 9.13783344 7.61820432 6.53327433 5.72003059 5.08791031 4.58257569 4.16944811 3.82547504 3.53469699 3.28571429 3.07016708 2.88178547 2.71577476 2.5684057 2.43673411 2.31840462 2.21151046 2.11449152 2.02605897 1.94513876 1.87082869 1.80236531 1.73909825 1.68047005 1.6260001 1.57527188 1.52792256 1.48363468 1.44212925 1.40316005 1.36650903 1.33198234 1.29940707 1.2686285 1.2395078 1.21191996 1.18575222 1.16090251 1.13727826 1.11479532 1.09337699 1.07295328 1.0534601

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392

Appendix F Table for Fanno Line Flow

(continued) γ = 1.4 Mach Number

fL*/Dh

p/p*

pt/pt*

T/T*

ρ/ρ*

0.96 0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24 1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42 1.44 1.46 1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68 1.7 1.72 1.74 1.76 1.78 1.8 1.82 1.84 1.86 1.88

0.00205714 0.0004947 0 0.00045869 0.0017685 0.00383785 0.0065846 0.009935 0.01382273 0.01818811 0.02297735 0.02814193 0.03363807 0.03942619 0.04547052 0.05173869 0.05820139 0.06483209 0.07160673 0.07850352 0.0855027 0.09258633 0.09973817 0.10694349 0.11418891 0.12146232 0.12875273 0.13605022 0.14334576 0.15063121 0.15789921 0.1651431 0.17235689 0.17953517 0.18667309 0.19376628 0.20081083 0.20780326 0.21474046 0.22161967 0.22843844 0.23519464 0.24188637 0.24851201 0.25507015 0.26155959 0.2679793

1.04853965 1.02379227 1 0.97710808 0.95506595 0.93382684 0.91334746 0.89358761 0.87450999 0.85607987 0.83826492 0.82103495 0.80436182 0.78821916 0.77258232 0.7574282 0.7427351 0.72848265 0.71465171 0.70122424 0.68818327 0.67551278 0.66319764 0.65122357 0.63957707 0.62824535 0.6172163 0.60647843 0.59602087 0.58583326 0.57590577 0.56622907 0.55679425 0.54759286 0.53861681 0.52985842 0.52131033 0.51296555 0.50481738 0.4968594 0.48908551 0.48148984 0.47406677 0.46681093 0.45971718 0.45278055 0.44599632

1.00136456 1.00033714 1 1.00032972 1.00130519 1.00290738 1.00511921 1.00792534 1.01131206 1.01526712 1.01977964 1.02484001 1.03043975 1.0365715 1.04322887 1.05040642 1.05809955 1.06630452 1.07501831 1.08423862 1.09396385 1.10419301 1.11492571 1.12616214 1.13790302 1.15014957 1.16290352 1.17616705 1.18994279 1.2042338 1.21904353 1.23437585 1.250235 1.26662557 1.28355253 1.30102118 1.31903715 1.3376064 1.3567352 1.37643014 1.3966981 1.41754625 1.43898206 1.46101327 1.48364792 1.5068943 1.53076097

1.01323966 1.00664385 1 0.9933117 0.98658248 0.97981579 0.97301505 0.96618357 0.95932464 0.95244142 0.94553707 0.9386146 0.93167702 0.92472721 0.91776799 0.91080211 0.90383225 0.89686099 0.88989084 0.88292425 0.87596356 0.86901107 0.86206897 0.85513939 0.84822438 0.84132593 0.83444593 0.82758621 0.82074852 0.81393456 0.80714593 0.80038418 0.79365079 0.78694717 0.78027466 0.77363454 0.76702802 0.76045627 0.75392039 0.7474214 0.74096028 0.73453798 0.72815534 0.72181319 0.71551231 0.70925339 0.70303712

1.03483873 1.01703524 1 0.98368727 0.96805484 0.95306367 0.93867762 0.92486318 0.91158922 0.89882679 0.88654897 0.87473064 0.86334835 0.8523802 0.8418057 0.83160567 0.82176211 0.81225816 0.80307795 0.79420658 0.78563002 0.77733507 0.76930926 0.76154084 0.75401873 0.74673242 0.73967201 0.73282811 0.72619183 0.71975474 0.71350886 0.7074466 0.70156076 0.6958445 0.6902913 0.68489499 0.67964966 0.6745497 0.66958977 0.66476476 0.66006981 0.65550026 0.6510517 0.64671987 0.64250073 0.6383904 0.63438517

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393

Appendix F Table for Fanno Line Flow (continued) γ = 1.4 Mach Number

fL*/Dh

p/p*

pt/pt*

T/T*

ρ/ρ*

1.9 1.92 1.94 1.96 1.98 2 2.02 2.04 2.06 2.08 2.1 2.12 2.14 2.16 2.18 2.2 2.22 2.24 2.26 2.28 2.3 2.32 2.34 2.36 2.38 2.4 2.42 2.44 2.46 2.48 2.5 2.52 2.54 2.56 2.58 2.6 2.62 2.64 2.66 2.68 2.7 2.72 2.74 2.76 2.78 2.8 2.82

0.27432844 0.28060634 0.28681244 0.29294634 0.29900775 0.3049965 0.31091251 0.31675578 0.32252642 0.32822461 0.33385058 0.33940465 0.34488716 0.35029855 0.35563927 0.36090982 0.36611074 0.37124261 0.37630603 0.38130164 0.38623008 0.39109204 0.39588821 0.40061929 0.40528602 0.40988913 0.41442936 0.41890747 0.42332422 0.42768037 0.43197669 0.43621396 0.44039294 0.44451442 0.44857915 0.45258792 0.4565415 0.46044064 0.46428612 0.46807868 0.47181909 0.47550808 0.47914641 0.48273481 0.48627401 0.48976473 0.49320769

0.43935993 0.43286701 0.42651334 0.4202949 0.4142078 0.40824829 0.40241278 0.3966978 0.39110002 0.3856162 0.38024325 0.37497818 0.3698181 0.36476022 0.35980184 0.35494037 0.35017329 0.34549818 0.34091269 0.33641454 0.33200154 0.32767157 0.32342256 0.31925253 0.31515954 0.31114172 0.30719726 0.30332441 0.29952146 0.29578675 0.2921187 0.28851573 0.28497635 0.28149909 0.27808252 0.27472527 0.271426 0.26818339 0.26499619 0.26186316 0.25878311 0.25575486 0.2527773 0.24984932 0.24696984 0.24413783 0.24135227

1.55525678 1.5803908 1.6061724 1.63261118 1.65971701 1.6875 1.71597051 1.74513915 1.77501677 1.80561448 1.83694361 1.86901575 1.90184273 1.93543661 1.9698097 2.00497455 2.04094392 2.07773086 2.11534861 2.15381069 2.19313082 2.23332298 2.27440139 2.31638051 2.35927503 2.40309988 2.44787024 2.49360153 2.5403094 2.58800976 2.63671875 2.68645276 2.73722843 2.78906263 2.84197249 2.89597538 2.95108894 3.00733102 3.06471976 3.12327353 3.18301096 3.24395092 3.30611258 3.3695153 3.43417876 3.50012286 3.56736777

0.69686411 0.69073494 0.68465014 0.67861021 0.67261558 0.66666667 0.66076384 0.65490744 0.64909775 0.64333505 0.63761955 0.63195147 0.62633095 0.62075815 0.61523317 0.6097561 0.60432698 0.59894586 0.59361273 0.58832758 0.58309038 0.57790106 0.57275956 0.56766576 0.56261956 0.55762082 0.55266939 0.54776512 0.54290781 0.53809729 0.53333333 0.52861573 0.52394425 0.51931865 0.51473868 0.51020408 0.50571457 0.50126988 0.49686972 0.49251379 0.48820179 0.48393341 0.47970834 0.47552625 0.47138682 0.46728972 0.46323461

0.6304815 0.62667599 0.62296539 0.61934657 0.61581654 0.61237244 0.6090115 0.6057311 0.60252868 0.59940182 0.59634817 0.59336548 0.59045158 0.58760439 0.58482191 0.5821022 0.57944342 0.57684376 0.57430151 0.57181501 0.56938265 0.56700288 0.56467423 0.56239526 0.56016456 0.55798082 0.55584273 0.55374904 0.55169856 0.5496901 0.54772256 0.54579483 0.54390587 0.54205465 0.54024019 0.53846154 0.53671777 0.53500799 0.53333133 0.53168696 0.53007406 0.52849185 0.52693956 0.52541646 0.52392182 0.52245496 0.5210152

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394

Appendix F Table for Fanno Line Flow

(continued) γ = 1.4 Mach Number

fL*/Dh

p/p*

pt/pt*

T/T*

ρ/ρ*

2.84 2.86 2.88 2.9 2.92 2.94 2.96 2.98 3 3.02 3.04 3.06 3.08 3.1 3.12 3.14 3.16 3.18 3.2 3.22 3.24 3.26 3.28 3.3 3.32 3.34 3.36 3.38 3.4 3.42 3.44 3.46 3.48 3.5 3.52 3.54 3.56 3.58 3.6 3.62 3.64 3.66 3.68 3.7 3.72 3.74 3.76

0.4966036 0.49995315 0.50325705 0.50651597 0.5097306 0.5129016 0.51602963 0.51911535 0.52215941 0.52516243 0.52812504 0.53104787 0.53393152 0.53677659 0.53958369 0.54235338 0.54508625 0.54778286 0.55044378 0.55306956 0.55566073 0.55821783 0.56074138 0.56323191 0.56568992 0.56811592 0.5705104 0.57287384 0.57520672 0.57750952 0.57978269 0.58202669 0.58424197 0.58642898 0.58858813 0.59071987 0.59282461 0.59490275 0.59695472 0.59898091 0.6009817 0.60295749 0.60490866 0.60683557 0.60873861 0.61061812 0.61247446

0.23861218 0.23591659 0.23326456 0.23065519 0.22808759 0.22556089 0.22307425 0.22062685 0.21821789 0.21584659 0.21351218 0.21121393 0.20895111 0.20672301 0.20452896 0.20236827 0.20024029 0.19814439 0.19607994 0.19404633 0.19204296 0.19006926 0.18812466 0.1862086 0.18432055 0.18245998 0.18062636 0.17881921 0.17703802 0.17528231 0.17355162 0.17184549 0.17016346 0.16850509 0.16686997 0.16525766 0.16366775 0.16209985 0.16055357 0.15902851 0.1575243 0.15604058 0.15457699 0.15313316 0.15170877 0.15030346 0.14891692

3.63593394 3.70584205 3.77711307 3.84976823 3.92382903 3.99931722 4.07625485 4.15466422 4.2345679 4.31598875 4.39894991 4.48347476 4.56958701 4.6573106 4.7466698 4.83768914 4.93039341 5.02480774 5.1209575 5.21886838 5.31856634 5.42007765 5.52342887 5.62864684 5.73575872 5.84479196 5.95577431 6.06873381 6.18369882 6.30069801 6.41976035 6.5409151 6.66419187 6.78962054 6.91723133 7.04705477 7.1791217 7.31346329 7.45011103 7.58909671 7.73045248 7.87421078 8.0204044 8.16906645 8.32023037 8.47392994 8.63019926

0.45922116 0.45524902 0.45131785 0.44742729 0.443577 0.43976663 0.43599581 0.4322642 0.42857143 0.42491714 0.42130098 0.41772258 0.41418158 0.41067762 0.40721034 0.40377937 0.40038437 0.39702496 0.39370079 0.39041149 0.38715672 0.38393611 0.38074931 0.37759597 0.37447573 0.37138825 0.36833317 0.36531015 0.36231884 0.3593589 0.35643 0.35353178 0.35066392 0.34782609 0.34501794 0.34223916 0.33948941 0.33676837 0.33407572 0.33141115 0.32877433 0.32616495 0.32358271 0.32102729 0.31849839 0.3159957 0.31351894

0.51960188 0.51821438 0.51685207 0.51551436 0.51420067 0.51291043 0.51164311 0.51039816 0.50917508 0.50797335 0.5067925 0.50563206 0.50449155 0.50337053 0.50226858 0.50118526 0.50012016 0.49907288 0.49804305 0.49703026 0.49603416 0.49505439 0.4940906 0.49314244 0.4922096 0.49129173 0.49038853 0.4894997 0.48862493 0.48776394 0.48691644 0.48608215 0.4852608 0.48445214 0.48365591 0.48287185 0.48209973 0.4813393 0.48059034 0.47985262 0.47912592 0.47841002 0.47770472 0.4770098 0.47632507 0.47565034 0.47498541

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395

Appendix F Table for Fanno Line Flow (continued) γ = 1.4 Mach Number

fL*/Dh

p/p*

pt/pt*

T/T*

ρ/ρ*

3.78 3.8 3.82 3.84 3.86 3.88 3.9 3.92 3.94 3.96 3.98 4 4.02 4.04 4.06 4.08 4.1 4.12 4.14 4.16 4.18 4.2 4.22 4.24 4.26 4.28 4.3 4.32 4.34 4.36 4.38 4.4 4.42 4.44 4.46 4.48 4.5 4.52 4.54 4.56 4.58 4.6 4.62 4.64 4.66 4.68 4.7

0.61430798 0.61611903 0.61790794 0.61967505 0.62142067 0.62314514 0.62484877 0.62653186 0.62819472 0.62983766 0.63146097 0.63306493 0.63464984 0.63621597 0.6377636 0.639293 0.64080443 0.64229817 0.64377446 0.64523356 0.64667572 0.64810119 0.6495102 0.65090299 0.65227981 0.65364086 0.65498639 0.65631661 0.65763174 0.658932 0.66021759 0.66148873 0.66274562 0.66398845 0.66521743 0.66643275 0.6676346 0.66882318 0.66999865 0.67116122 0.67231104 0.67344831 0.6745732 0.67568587 0.67678649 0.67787523 0.67895225

0.14754881 0.14619883 0.14486666 0.143552 0.14225454 0.14097401 0.13971012 0.13846258 0.13723113 0.13601549 0.13481541 0.13363062 0.13246088 0.13130594 0.13016555 0.12903949 0.12792751 0.12682939 0.12574491 0.12467384 0.12361598 0.12257111 0.12153903 0.12051953 0.11951242 0.1185175 0.11753458 0.11656347 0.115604 0.11465597 0.11371922 0.11279356 0.11187883 0.11097487 0.11008151 0.10919858 0.10832593 0.1074634 0.10661085 0.10576812 0.10493507 0.10411155 0.10329742 0.10249255 0.10169679 0.10091001 0.10013209

8.78907277 8.95058526 9.11477185 9.28166799 9.45130949 9.62373249 9.79897348 9.97706931 10.1580572 10.3419746 10.5288594 10.71875 10.9116849 11.107703 11.3068437 11.5091467 11.7146519 11.9233999 12.1354314 12.3507874 12.5695097 12.7916399 13.0172205 13.2462939 13.4789033 13.7150919 13.9549037 14.1983827 14.4455734 14.6965208 14.9512702 15.2098673 15.4723581 15.7387892 16.0092074 16.28366 16.5621946 16.8448593 17.1317026 17.4227734 17.7181209 18.0177948 18.3218452 18.6303226 18.9432778 19.2607624 19.5828278

0.31106779 0.30864198 0.3062412 0.30386516 0.3015136 0.29918621 0.29688273 0.29460288 0.29234637 0.29011295 0.28790234 0.28571429 0.28354852 0.28140477 0.2792828 0.27718235 0.27510316 0.273045 0.27100761 0.26899075 0.26699418 0.26501767 0.26306098 0.26112388 0.25920614 0.25730753 0.25542784 0.25356684 0.25172431 0.24990004 0.24809381 0.24630542 0.24453465 0.2427813 0.24104517 0.23932606 0.23762376 0.23593809 0.23426885 0.23261584 0.23097889 0.2293578 0.22775239 0.22616248 0.22458788 0.22302843 0.22148394

0.47433009 0.47368421 0.47304759 0.47242005 0.47180142 0.47119154 0.47059025 0.46999739 0.4694128 0.46883633 0.46826784 0.46770717 0.4671542 0.46660878 0.46607077 0.46554005 0.46501649 0.46449995 0.46399032 0.46348747 0.46299129 0.46250166 0.46201847 0.4615416 0.46107095 0.46060641 0.46014788 0.45969526 0.45924844 0.45880733 0.45837183 0.45794186 0.45751731 0.45709809 0.45668413 0.45627533 0.45587162 0.45547289 0.45507909 0.45469011 0.4543059 0.45392637 0.45355145 0.45318105 0.45281512 0.45245359 0.45209637

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396

Appendix F Table for Fanno Line Flow

(continued) γ = 1.4 Mach Number

fL*/Dh

p/p*

pt/pt*

T/T*

ρ/ρ*

4.72 4.74 4.76 4.78 4.8 4.82 4.84 4.86 4.88 4.9 4.92 4.94 4.96 4.98 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 12 14 16 18 20

0.68001771 0.68107177 0.68211457 0.68314628 0.68416705 0.68517701 0.68617633 0.68716513 0.68814355 0.68911175 0.69006985 0.69101799 0.69195629 0.69288489 0.69380392 0.71400429 0.72987528 0.74254432 0.75280216 0.76121412 0.76819161 0.77403906 0.77898519 0.78320428 0.78683083 0.79721157 0.80356079 0.80771861 0.81058621 0.81264596

0.09936288 0.09860227 0.09785013 0.09710634 0.09637078 0.09564333 0.09492387 0.09421229 0.09350848 0.09281233 0.09212373 0.09144257 0.09076875 0.09010217 0.08944272 0.07501245 0.06375767 0.05482282 0.04761905 0.04173124 0.03686049 0.03278744 0.02934836 0.02641919 0.02390457 0.0167225 0.01234098 0.00947623 0.00750249 0.00608581

19.9095265 20.2409108 20.5770338 20.9179489 21.2637099 21.6143712 21.9699874 22.3306137 22.6963055 23.067119 23.4431104 23.8243367 24.2108552 24.6027235 25 36.8689631 53.179784 75.1343149 104.142857 141.841483 190.109375 251.086167 327.1893 421.131373 535.9375 1276.21489 2685.38393 5144.55469 9159.28215 15377.3438

0.21995425 0.21843918 0.21693856 0.21545223 0.21398003 0.21252178 0.21107734 0.20964654 0.20822922 0.20682523 0.20543443 0.20405665 0.20269175 0.20133958 0.2 0.17021277 0.14634146 0.12698413 0.11111111 0.09795918 0.08695652 0.0776699 0.06976744 0.06299213 0.05714286 0.04026846 0.02985075 0.02298851 0.01823708 0.01481481

0.45174341 0.45139463 0.45104998 0.45070939 0.45037279 0.45004012 0.44971133 0.44938636 0.44906514 0.44874762 0.44843375 0.44812346 0.44781671 0.44751344 0.4472136 0.44069817 0.43567742 0.4317297 0.42857143 0.42600643 0.42389562 0.42213824 0.42065988 0.41940467 0.41833001 0.41527546 0.41342275 0.41221581 0.41138629 0.41079192

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APPENDIX G

Table for Rayleigh Line Flow

Note – The numerical values of the table can be directly obtained from the following equations. They are tabulated for γ ¼ 1:4 for convenience. (The subscripts t for thermodynamic properties denote stagnation values; the superscript * denotes critical condition for M ¼ 1.)

pt ¼ pt



p 1þγ ¼ p 1 þ γM2

(7.10)

T ð1 þ γÞ2 M2 ¼ 2  T ð1 þ γM2 Þ

(7.11)

V ρ p =RT  ð1 þ γÞM2 ¼ ¼ ¼  V ρ p=RT 1 þ γM2

(7.12)

1þγ 1 þ γM2



2 γþ1



 γ  1 2 γ=ðγ M 1þ 2



Tt 2ð1 þ γÞM2 γ 1 2 M ¼ 1þ 2 Tt ð1 þ γM2 Þ2

1Þ

(7.13)

(7.14)

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398

Appendix G Table for Rayleigh Line Flow

γ = 1.4 Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9 0.92 0.94

2.4 2.39865675 2.39463602 2.38796466 2.37868696 2.36686391 2.35257215 2.33590283 2.31696015 2.2958598 2.27272727 2.24769611 2.22090613 2.19250164 2.16262976 2.13143872 2.09907641 2.06568891 2.03141928 1.99640647 1.96078431 1.92468082 1.88821752 1.85150898 1.81466247 1.77777778 1.74094708 1.70425496 1.66777852 1.63158753 1.59574468 1.56030582 1.52532032 1.49083139 1.45687646 1.42348754 1.39069164 1.35851107 1.32696391 1.29606428 1.26582278 1.23624675 1.20734063 1.17910624 1.15154307 1.12464855 1.09841828 1.07284626

0 0.00230142 0.00917485 0.02052855 0.03621217 0.05602045 0.07969818 0.10694626 0.13742859 0.1707795 0.20661157 0.24452347 0.28410762 0.32495749 0.36667425 0.40887279 0.45118687 0.49327337 0.53481569 0.57552624 0.61514802 0.6534555 0.69025474 0.72538289 0.75870717 0.79012346 0.81955447 0.84694781 0.87227376 0.895523 0.91670439 0.93584265 0.95297621 0.96815507 0.98143892 0.99289523 1.00259764 1.01062446 1.01705726 1.02197975 1.02547669 1.02763298 1.02853294 1.02825961 1.02689423 1.02451583 1.02120082 1.0170228

0 0.00095946 0.00383142 0.00859667 0.0152236 0.02366864 0.03387704 0.0457837 0.05931418 0.07438586 0.09090909 0.10878849 0.12792419 0.14821311 0.16955017 0.19182948 0.21494542 0.23879364 0.26327194 0.28828109 0.31372549 0.3395137 0.36555891 0.3917793 0.41809823 0.44444444 0.47075209 0.49696075 0.52301534 0.54886605 0.57446809 0.59978156 0.6247712 0.64940615 0.67365967 0.6975089 0.72093454 0.74392066 0.76645435 0.78852551 0.81012658 0.83125232 0.85189955 0.87206697 0.89175495 0.91096532 0.92970123 0.94796696

1.26787629 1.26752152 1.26646001 1.26470013 1.26225566 1.2591456 1.25539382 1.25102873 1.24608281 1.24059214 1.23459588 1.22813574 1.22125541 1.21400003 1.20641566 1.19854878 1.1904458 1.18215267 1.17371444 1.16517497 1.15657661 1.14795997 1.13936373 1.1308245 1.1223767 1.1140525 1.10588181 1.09789224 1.0901092 1.08255586 1.07525332 1.06822062 1.06147487 1.05503135 1.04890365 1.04310374 1.03764211 1.0325279 1.027769 1.02337216 1.01934312 1.01568668 1.01240683 1.00950681 1.00698925 1.00485619 1.0031092 1.00174943

0 0.001918 0.00764816 0.01711944 0.03021544 0.04677707 0.06660642 0.08947124 0.11511019 0.14323846 0.17355372 0.20574205 0.23948379 0.2744591 0.31035308 0.34686042 0.38368931 0.42056487 0.45723176 0.4934562 0.5290273 0.56375784 0.59748451 0.63006758 0.66139033 0.69135802 0.71989665 0.74695151 0.77248564 0.79647816 0.81892259 0.8398252 0.85920335 0.87708395 0.89350199 0.90849913 0.92212247 0.93442337 0.94545643 0.95527854 0.96394809 0.97152422 0.97806626 0.98363314 0.98828301 0.99207283 0.99505808 0.99729256

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399

Appendix G Table for Rayleigh Line Flow (continued) γ = 1.4 Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

0.96 0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24 1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42 1.44 1.46 1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68 1.7 1.72 1.74 1.76 1.78 1.8 1.82 1.84 1.86 1.88

1.04792511 1.02364623 1 0.97697593 0.95456281 0.93274881 0.91152163 0.8908686 0.87077673 0.85123287 0.8322237 0.81373586 0.79575597 0.77827068 0.76126675 0.74473103 0.72865054 0.71301248 0.69780424 0.68301346 0.66862798 0.65463591 0.64102564 0.6277858 0.6149053 0.60237335 0.59017941 0.57831325 0.56676491 0.55552469 0.54458321 0.53393134 0.52356021 0.51346124 0.50362611 0.49404674 0.48471531 0.47562426 0.46676624 0.45813417 0.44972117 0.4415206 0.43352601 0.42573119 0.41813012 0.41071697 0.40348612

1.01205231 1.00635668 1 0.99304305 0.98554328 0.97755486 0.96912874 0.9603127 0.95115146 0.94168678 0.93195757 0.92200001 0.91184769 0.9015317 0.89108081 0.88052155 0.86987835 0.85917368 0.84842816 0.83766065 0.82688841 0.81612715 0.80539119 0.79469351 0.78404586 0.77345885 0.76294204 0.75250399 0.74215237 0.73189398 0.72173488 0.71168038 0.70173515 0.69190324 0.68218814 0.67259285 0.66311987 0.65377127 0.64454876 0.63545364 0.62648692 0.61764928 0.60894116 0.60036273 0.59191392 0.5835945 0.57540403

0.96576778 0.98310984 1 1.01644576 1.03245514 1.04803656 1.06319883 1.077951 1.09230233 1.10626224 1.11984021 1.13304581 1.14588859 1.15837808 1.17052375 1.18233498 1.19382104 1.20499109 1.21585411 1.22641896 1.2366943 1.24668863 1.25641026 1.26586729 1.27506764 1.28401903 1.29272899 1.30120482 1.30945364 1.31748236 1.32529771 1.33290619 1.34031414 1.34752768 1.35455278 1.36139519 1.36806049 1.3745541 1.38088125 1.38704702 1.39305631 1.39891386 1.40462428 1.410192 1.41562134 1.42091645 1.42608134

1.00077767 1.00019444 1 1.00019444 1.00077769 1.0017496 1.00310993 1.00485842 1.00699479 1.00951882 1.01243029 1.01572906 1.0194151 1.02348846 1.02794929 1.0327979 1.03803471 1.04366031 1.04967542 1.05608095 1.06287797 1.0700677 1.07765156 1.08563116 1.09400827 1.10278485 1.11196306 1.12154523 1.13153389 1.14193177 1.15274177 1.163967 1.17561073 1.18767645 1.20016783 1.21308872 1.22644318 1.24023542 1.25446987 1.26915114 1.28428402 1.29987347 1.31592466 1.33244292 1.34943378 1.36690294 1.38485627

0.99882816 0.99971472 1 0.99972954 0.99894666 0.99769249 0.99600591 0.99392364 0.99148028 0.98870834 0.98563832 0.98229881 0.97871652 0.97491638 0.97092165 0.96675396 0.9624334 0.95797865 0.953407 0.94873445 0.94397581 0.93914472 0.93425378 0.92931459 0.92433779 0.91933319 0.91430974 0.90927566 0.90423845 0.89920495 0.8941814 0.88917347 0.88418629 0.87922451 0.87429233 0.86939352 0.86453147 0.85970922 0.85492947 0.8501946 0.84550674 0.84086773 0.83627919 0.83174252 0.8272589 0.82282935 0.81845469

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400

Appendix G Table for Rayleigh Line Flow

(continued) γ = 1.4 Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

1.9 1.92 1.94 1.96 1.98 2 2.02 2.04 2.06 2.08 2.1 2.12 2.14 2.16 2.18 2.2 2.22 2.24 2.26 2.28 2.3 2.32 2.34 2.36 2.38 2.4 2.42 2.44 2.46 2.48 2.5 2.52 2.54 2.56 2.58 2.6 2.62 2.64 2.66 2.68 2.7 2.72 2.74 2.76 2.78 2.8 2.82

0.39643211 0.38954968 0.38283374 0.37627935 0.36988176 0.36363636 0.3575387 0.35158447 0.34576951 0.34008978 0.3345414 0.32912059 0.32382371 0.31864724 0.31358776 0.30864198 0.3038067 0.29907884 0.2944554 0.28993351 0.28551035 0.28118322 0.27694949 0.27280663 0.26875218 0.26478376 0.26089906 0.25709585 0.25337196 0.2497253 0.24615385 0.24265562 0.23922873 0.23587131 0.23258158 0.2293578 0.22619829 0.22310141 0.22006558 0.21708927 0.21417098 0.21130927 0.20850274 0.20575003 0.20304981 0.2004008 0.19780176

0.56734189 0.55940734 0.5515995 0.54391735 0.53635979 0.52892562 0.52161355 0.51442221 0.50735019 0.50039601 0.49355815 0.48683504 0.48022508 0.47372665 0.4673381 0.46105777 0.45488398 0.44881504 0.44284928 0.43698499 0.43122049 0.42555409 0.41998412 0.4145089 0.40912679 0.40383613 0.3986353 0.39352268 0.38849668 0.38355571 0.37869822 0.37392267 0.36922754 0.36461133 0.36007257 0.3556098 0.35122159 0.34690653 0.34266324 0.33849035 0.33438653 0.33035046 0.32638085 0.32247642 0.31863594 0.31485817 0.31114192

1.43111992 1.43603594 1.44083305 1.44551475 1.45008446 1.45454545 1.45890093 1.46315395 1.46730749 1.47136444 1.47532757 1.47919958 1.48298306 1.48668055 1.49029446 1.49382716 1.49728093 1.50065797 1.50396043 1.50719035 1.51034975 1.51344056 1.51646465 1.51942383 1.52231987 1.52515446 1.52792924 1.53064582 1.53330574 1.5359105 1.53846154 1.54096027 1.54340805 1.54580621 1.54815602 1.55045872 1.55271551 1.55492757 1.55709601 1.55922195 1.56130644 1.56335052 1.56535518 1.56732141 1.56925014 1.57114228 1.57299874

1.40329985 1.42223991 1.44168287 1.46163532 1.48210404 1.50309598 1.52461825 1.54667816 1.56928319 1.59244097 1.61615933 1.64044628 1.66530997 1.69075875 1.71680114 1.74344583 1.77070168 1.79857772 1.82708317 1.85622741 1.88601999 1.91647064 1.94758927 1.97938595 2.01187094 2.04505465 2.07894769 2.11356084 2.14890504 2.18499142 2.22183129 2.25943612 2.29781757 2.33698749 2.37695787 2.41774092 2.459349 2.50179467 2.54509067 2.58924991 2.63428548 2.68021066 2.72703893 2.77478392 2.82345948 2.87307962 2.92365854

0.81413561 0.80987266 0.80566623 0.80151661 0.79742398 0.79338843 0.78940994 0.78548842 0.78162371 0.77781556 0.7740637 0.77036776 0.76672736 0.76314205 0.75961134 0.75613474 0.75271168 0.74934159 0.74602389 0.74275795 0.73954313 0.73637879 0.73326427 0.73019888 0.72718195 0.7242128 0.72129071 0.71841501 0.71558498 0.71279994 0.71005917 0.70736199 0.70470769 0.70209558 0.69952498 0.6969952 0.69450556 0.6920554 0.68964403 0.68727081 0.68493508 0.68263619 0.68037352 0.67814642 0.67595428 0.67379649 0.67167244

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401

Appendix G Table for Rayleigh Line Flow (continued) γ = 1.4 Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

2.84 2.86 2.88 2.9 2.92 2.94 2.96 2.98 3 3.02 3.04 3.06 3.08 3.1 3.12 3.14 3.16 3.18 3.2 3.22 3.24 3.26 3.28 3.3 3.32 3.34 3.36 3.38 3.4 3.42 3.44 3.46 3.48 3.5 3.52 3.54 3.56 3.58 3.6 3.62 3.64 3.66 3.68 3.7 3.72 3.74 3.76

0.19525148 0.19274879 0.19029254 0.18788163 0.18551499 0.18319156 0.18091034 0.17867034 0.17647059 0.17431017 0.17218817 0.17010371 0.16805593 0.166044 0.16406711 0.16212448 0.16021533 0.15833892 0.15649452 0.15468143 0.15289896 0.15114645 0.14942323 0.14772867 0.14606216 0.1444231 0.1428109 0.14122499 0.1396648 0.13812981 0.13661949 0.13513331 0.13367078 0.1322314 0.13081471 0.12942024 0.12804753 0.12669614 0.12536565 0.12405563 0.12276566 0.12149536 0.12024434 0.1190122 0.11779858 0.11660312 0.11542546

0.30748601 0.30388929 0.30035061 0.29686887 0.29344297 0.29007186 0.28675448 0.2834898 0.28027682 0.27711455 0.27400203 0.2709383 0.26792245 0.26495357 0.26203075 0.25915315 0.25631988 0.25353014 0.25078308 0.24807792 0.24541387 0.24279016 0.24020603 0.23766075 0.2351536 0.23268387 0.23025086 0.22785391 0.22549234 0.22316551 0.22087278 0.21861352 0.21638714 0.21419302 0.21203059 0.20989927 0.2077985 0.20572774 0.20368644 0.20167408 0.19969013 0.19773411 0.19580551 0.19390384 0.19202863 0.19017943 0.18835576

1.57482037 1.57660801 1.57836247 1.58008455 1.58177501 1.5834346 1.58506404 1.58666405 1.58823529 1.58977845 1.59129417 1.59278307 1.59424576 1.59568286 1.59709492 1.59848251 1.59984619 1.60118649 1.60250391 1.60379898 1.60507217 1.60632397 1.60755484 1.60876523 1.6099556 1.61112636 1.61227793 1.61341072 1.61452514 1.61562156 1.61670037 1.61776192 1.61880659 1.61983471 1.62084663 1.62184269 1.62282319 1.62378847 1.62473882 1.62567455 1.62659595 1.62750331 1.6283969 1.629277 1.63014387 1.63099777 1.63183896

2.97521066 3.02775053 3.08129294 3.13585286 3.19144542 3.24808598 3.30579008 3.36457344 3.42445199 3.48544186 3.54755935 3.61082099 3.67524349 3.74084377 3.80763893 3.87564629 3.94488336 4.01536788 4.08711775 4.16015112 4.23448631 4.31014188 4.38713656 4.46548934 4.54521937 4.62634604 4.70888894 4.7928679 4.87830292 4.96521426 5.05362237 5.14354793 5.23501183 5.32803518 5.42263933 5.51884583 5.61667647 5.71615326 5.81729842 5.92013442 6.02468394 6.13096992 6.23901548 6.34884402 6.46047915 6.5739447 6.68926478

0.66958154 0.66752321 0.66549685 0.66350192 0.66153784 0.65960407 0.65770007 0.6558253 0.65397924 0.65216138 0.65037121 0.64860824 0.64687197 0.64516194 0.64347766 0.64181868 0.64018454 0.63857481 0.63698904 0.6354268 0.63388767 0.63237125 0.63087712 0.62940489 0.62795417 0.62652458 0.62511574 0.62372728 0.62235885 0.62101009 0.61968066 0.61837021 0.61707841 0.61580493 0.61454946 0.61331167 0.61209127 0.61088795 0.6097014 0.60853135 0.60737751 0.60623959 0.60511733 0.60401046 0.6029187 0.60184181 0.60077953

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402

Appendix G Table for Rayleigh Line Flow

(continued) γ = 1.4 Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

3.78 3.8 3.82 3.84 3.86 3.88 3.9 3.92 3.94 3.96 3.98 4 4.02 4.04 4.06 4.08 4.1 4.12 4.14 4.16 4.18 4.2 4.22 4.24 4.26 4.28 4.3 4.32 4.34 4.36 4.38 4.4 4.42 4.44 4.46 4.48 4.5 4.52 4.54 4.56 4.58 4.6 4.62 4.64 4.66 4.68 4.7

0.11426526 0.11312217 0.11199588 0.11088605 0.10979238 0.10871456 0.10765228 0.10660526 0.10557321 0.10455585 0.1035529 0.1025641 0.10158919 0.10062792 0.09968003 0.09874528 0.09782343 0.09691425 0.09601751 0.095133 0.09426048 0.09339975 0.0925506 0.09171283 0.09088623 0.09007062 0.08926579 0.08847157 0.08768776 0.0869142 0.0861507 0.0853971 0.08465322 0.0839189 0.08319398 0.08247831 0.08177172 0.08107407 0.08038521 0.07970499 0.07903327 0.07836991 0.07771477 0.07706773 0.07642864 0.07579739 0.07517384

0.18655719 0.18478328 0.18303359 0.18130772 0.17960525 0.17792578 0.1762689 0.17463425 0.17302143 0.17143008 0.16985983 0.16831032 0.16678121 0.16527216 0.16378283 0.16231288 0.160862 0.15942988 0.15801619 0.15662065 0.15524294 0.15388278 0.15253988 0.15121396 0.14990475 0.14861197 0.14733537 0.14607467 0.14482963 0.1436 0.14238553 0.14118598 0.14000111 0.1388307 0.13767452 0.13653233 0.13540394 0.13428912 0.13318765 0.13209934 0.13102399 0.12996138 0.12891133 0.12787365 0.12684815 0.12583463 0.12483293

1.63266767 1.63348416 1.63428866 1.63508139 1.63586258 1.63663246 1.63739123 1.6381391 1.63887628 1.63960297 1.64031936 1.64102564 1.641722 1.64240863 1.64308569 1.64375337 1.64441184 1.64506125 1.64570178 1.64633357 1.6469568 1.64757161 1.64817814 1.64877655 1.64936698 1.64994956 1.65052444 1.65109174 1.6516516 1.65220414 1.6527495 1.65328779 1.65381913 1.65434364 1.65486144 1.65537264 1.65587734 1.65637566 1.65686771 1.65735358 1.65783338 1.65830721 1.65877516 1.65923734 1.65969383 1.66014472 1.66059011

6.80646368 6.92556597 7.04659644 7.16958013 7.2945423 7.42150848 7.55050442 7.68155612 7.81468983 7.94993206 8.08730953 8.22684925 8.36857846 8.51252465 8.65871556 8.80717919 8.95794381 9.11103792 9.26649028 9.42432994 9.58458616 9.7472885 9.91246677 10.080151 10.2503716 10.4231592 10.5985446 10.7765589 10.9572335 11.1406002 11.3266909 11.5155377 11.7071733 11.9016303 12.0989419 12.2991412 12.5022621 12.7083382 12.9174039 13.1294935 13.3446417 13.5628837 13.7842547 14.0087902 14.2365262 14.4674989 14.7017446

0.59973161 0.59869782 0.5976779 0.59667163 0.59567877 0.59469911 0.59373242 0.59277849 0.5918371 0.59090805 0.58999112 0.58908613 0.58819287 0.58731115 0.58644079 0.58558159 0.58473338 0.58389598 0.58306922 0.58225291 0.5814469 0.58065101 0.5798651 0.57908899 0.57832253 0.57756557 0.57681796 0.57607955 0.57535019 0.57462975 0.57391809 0.57321507 0.57252055 0.5718344 0.57115649 0.57048671 0.56982491 0.56917099 0.56852481 0.56788627 0.56725524 0.56663162 0.56601528 0.56540613 0.56480405 0.56420895 0.5636207

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403

Appendix G Table for Rayleigh Line Flow (continued) γ = 1.4 Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

4.72 4.74 4.76 4.78 4.8 4.82 4.84 4.86 4.88 4.9 4.92 4.94 4.96 4.98 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 12 14 16 18 20

0.07455787 0.07394936 0.0733482 0.07275426 0.07216743 0.0715876 0.07101466 0.0704485 0.06988902 0.06933611 0.06878967 0.0682496 0.06771581 0.0671882 0.06666667 0.05536332 0.04669261 0.03990025 0.03448276 0.03009404 0.02649007 0.02349486 0.02097902 0.0188457 0.01702128 0.011846 0.0087146 0.0066778 0.00527937 0.00427807

0.12384287 0.12286426 0.12189694 0.12094074 0.11999549 0.11906104 0.11813721 0.11722386 0.11632083 0.11542797 0.11454513 0.11367217 0.11280894 0.1119553 0.11111111 0.0927192 0.07848718 0.06726326 0.05826397 0.0509429 0.04491031 0.03988261 0.03564967 0.03205323 0.02897239 0.0202072 0.01488506 0.0114158 0.00903043 0.00732077

1.66103009 1.66146474 1.66189414 1.66231839 1.66273755 1.66315172 1.66356096 1.66396536 1.66436499 1.66475992 1.66515024 1.665536 1.66591728 1.66629415 1.66666667 1.67474048 1.68093385 1.68578554 1.68965517 1.69278997 1.69536424 1.69750367 1.6993007 1.7008245 1.70212766 1.70582428 1.708061 1.70951586 1.71051474 1.71122995

14.9393002 15.1802027 15.4244894 15.6721979 15.9233663 16.1780327 16.4362357 16.6980141 16.9634071 17.232454 17.5051946 17.781669 18.0619175 18.3459808 18.6338998 27.2113249 38.9459449 54.683031 75.4137931 102.287485 136.623525 179.923629 233.88395 300.407217 381.614879 904.053983 1896.28727 3625.3135 6445.22491 10809.6488

0.56303922 0.5624644 0.56189614 0.56133435 0.56077894 0.5602298 0.55968685 0.55915 0.55861916 0.55809424 0.55757516 0.55706184 0.55655418 0.55605211 0.55555556 0.54472528 0.53632909 0.5296982 0.52437574 0.52004206 0.51646858 0.51348863 0.51097853 0.50884502 0.50701675 0.50181208 0.49864962 0.49658724 0.49516881 0.49415196

Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

2.66666667 2.66489007 2.65957447 2.65076209 2.63852243 2.62295082 2.60416667 2.58231117 2.55754476 2.53004428 2.5

0 0.00284066 0.01131734 0.02529554 0.04455552 0.06879871 0.09765625 0.13069929 0.1674505 0.20739642 0.25

0 0.00106596 0.00425532 0.00954274 0.01688654 0.02622951 0.0375 0.0506133 0.06547315 0.08197343 0.1

1.29903811 1.29860542 1.29731135 1.29516772 1.292194 1.28841689 1.28386985 1.27859245 1.27262967 1.26603107 1.25885001

0 0.00213078 0.00849253 0.01899442 0.03348793 0.05177103 0.07359375 0.09866489 0.12665956 0.15722723 0.19

γ = 5/3

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404

Appendix G Table for Rayleigh Line Flow

(continued) γ = 5/3 Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9 0.92 0.94 0.96 0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14

2.46761258 2.43309002 2.3966447 2.35849057 2.31884058 2.27790433 2.23588597 2.19298246 2.14938205 2.10526316 2.06079341 2.01612903 1.97141449 1.92678227 1.88235294 1.83823529 1.79452669 1.75131349 1.70867151 1.66666667 1.62535555 1.58478605 1.54499807 1.5060241 1.46788991 1.43061516 1.39421401 1.35869565 1.32406488 1.29032258 1.25746621 1.2254902 1.19438638 1.16414435 1.13475177 1.10619469 1.07845781 1.05152471 1.02537811 1 0.97537186 0.95147479 0.92828963 0.9057971 0.8839779 0.86281277 0.84228259

0.29471301 0.3409878 0.38828803 0.43609826 0.48393195 0.53133805 0.57790591 0.6232687 0.66710536 0.70914127 0.74914777 0.78694069 0.82237813 0.85535768 0.88581315 0.91371107 0.93904708 0.96184222 0.98213942 1 1.01550048 1.02872958 1.03978549 1.04877341 1.05580338 1.06098841 1.06444279 1.06628072 1.06661512 1.06555671 1.06321317 1.05968858 1.05508291 1.04949172 1.04300589 1.03571149 1.02768975 1.01901701 1.00976481 1 0.98978482 0.9791771 0.96823042 0.95699433 0.94551448 0.93383291 0.92198816

0.11943245 0.14014599 0.16201318 0.18490566 0.20869565 0.2332574 0.25846842 0.28421053 0.31037077 0.33684211 0.36352396 0.39032258 0.41715131 0.44393064 0.47058824 0.49705882 0.52328398 0.54921191 0.5747971 0.6 0.62478667 0.64912837 0.67300116 0.69638554 0.71926606 0.7416309 0.76347159 0.78478261 0.80556107 0.82580645 0.84552028 0.86470588 0.88336817 0.90151339 0.91914894 0.93628319 0.95292532 0.96908517 0.98477314 1 1.01477688 1.02911513 1.04302622 1.05652174 1.06961326 1.08231234 1.09463045

1.2511428 1.24296779 1.23438461 1.22545337 1.21623392 1.20678522 1.19716472 1.18742794 1.17762796 1.16781519 1.15803703 1.14833776 1.13875839 1.12933664 1.12010692 1.11110041 1.10234511 1.09386598 1.08568511 1.0778218 1.07029283 1.06311255 1.05629311 1.04984465 1.04377544 1.03809208 1.03279965 1.02790188 1.02340132 1.01929942 1.01559672 1.01229294 1.00938708 1.00687757 1.0047623 1.00303874 1.00170401 1.00075493 1.00018812 1 1.00018687 1.00074495 1.00167041 1.0029594 1.00460809 1.00661265 1.00896936

0.22460079 0.26065107 0.29777809 0.33562122 0.37383743 0.41210579 0.45013091 0.48764543 0.52441152 0.56022161 0.59489825 0.62829344 0.6602874 0.69078686 0.71972318 0.74705017 0.77274184 0.7967901 0.81920249 0.84 0.85921496 0.8768891 0.89307176 0.90781826 0.92118845 0.93324541 0.94405431 0.95368147 0.9621935 0.96965661 0.97613602 0.9816955 0.98639702 0.99030039 0.99346311 0.99594017 0.99778397 0.99904427 0.99976814 1 0.99978164 0.99915231 0.99814874 0.99680529 0.99515399 0.99322468 0.99104508

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405

Appendix G Table for Rayleigh Line Flow (continued) γ = 5/3 Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

1.16 1.18 1.2 1.22 1.24 1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42 1.44 1.46 1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68 1.7 1.72 1.74 1.76 1.78 1.8 1.82 1.84 1.86 1.88 1.9 1.92 1.94 1.96 1.98 2 2.2 2.4 2.6 2.8

0.82236842 0.8030516 0.78431373 0.76613676 0.74850299 0.73139514 0.71479628 0.69868996 0.68306011 0.66789113 0.65316786 0.63887558 0.625 0.61152729 0.59844405 0.5857373 0.5733945 0.56140351 0.54975261 0.53843048 0.52742616 0.5167291 0.50632911 0.49621635 0.48638132 0.47681488 0.46750818 0.45845272 0.44964029 0.44106296 0.43271311 0.42458338 0.41666667 0.40895614 0.4014452 0.3941275 0.3869969 0.38004751 0.37327361 0.36666972 0.36023055 0.35395098 0.34782609 0.29411765 0.25157233 0.2173913 0.18957346

0.91001558 0.89794743 0.88581315 0.87363949 0.86145075 0.84926891 0.83711382 0.82500334 0.81295351 0.80097867 0.78909163 0.77730372 0.765625 0.75406429 0.74262931 0.73132677 0.72016244 0.70914127 0.69826742 0.68754433 0.67697484 0.66656119 0.65630508 0.64620776 0.63627004 0.62649235 0.61687475 0.60741702 0.59811863 0.58897882 0.5799966 0.57117078 0.5625 0.55398275 0.54561737 0.53740211 0.52933508 0.52141435 0.51363786 0.50600354 0.49850925 0.49115279 0.48393195 0.41868512 0.36454254 0.3194707 0.28175468

1.10657895 1.11816904 1.12941176 1.14031795 1.1508982 1.16116292 1.17112223 1.18078603 1.19016393 1.19926532 1.20809928 1.21667465 1.225 1.23308363 1.24093357 1.24855762 1.2559633 1.26315789 1.27014843 1.27694171 1.2835443 1.28996254 1.29620253 1.30227019 1.30817121 1.31391107 1.31949509 1.32492837 1.33021583 1.33536222 1.34037213 1.34524997 1.35 1.35462632 1.35913288 1.3635235 1.36780186 1.3719715 1.37603583 1.37999817 1.38386167 1.38762941 1.39130435 1.42352941 1.4490566 1.46956522 1.48625592

1.01167453 1.01472458 1.018116 1.02184542 1.02590958 1.03030534 1.03502968 1.04007972 1.04545273 1.05114609 1.05715734 1.06348414 1.07012431 1.07707578 1.08433664 1.09190509 1.09977948 1.10795827 1.11644006 1.12522356 1.13430762 1.14369118 1.15337332 1.16335321 1.17363013 1.18420347 1.19507272 1.20623747 1.21769739 1.22945227 1.24150195 1.25384638 1.26648559 1.2794197 1.29264887 1.30617337 1.31999353 1.33410975 1.34852249 1.36323228 1.37823971 1.39354543 1.40915015 1.58183529 1.78555468 2.02170011 2.29192691

0.98864093 0.98603608 0.9832526 0.98031087 0.97722973 0.97402651 0.97071718 0.96731641 0.96383768 0.96029333 0.95669469 0.95305209 0.949375 0.94567202 0.94195101 0.93821911 0.93448279 0.93074792 0.92701982 0.92330329 0.91960263 0.91592173 0.91226406 0.90863273 0.90503051 0.90145984 0.89792289 0.89442156 0.89095751 0.88753218 0.88414681 0.88080246 0.8775 0.87424018 0.87102358 0.86785067 0.86472179 0.86163721 0.85859705 0.85560139 0.85265022 0.84974344 0.84688091 0.82062284 0.79834817 0.77950851 0.76355518

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406

Appendix G Table for Rayleigh Line Flow

(continued) γ = 5/3 Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

3 3.2 3.4 3.6 3.8 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 12 14 16 18 20

0.16666667 0.14760148 0.13157895 0.1179941 0.10638298 0.09638554 0.07673861 0.0625 0.05186386 0.04371585 0.03733956 0.03225806 0.02814424 0.0247678 0.02196294 0.01960784 0.01761145 0.01590457 0.01106501 0.00813835 0.00623539 0.00492914 0.00399401

0.25 0.22309064 0.2001385 0.180437 0.16342236 0.14864276 0.11924849 0.09765625 0.08136826 0.06879871 0.05890675 0.05098855 0.04455552 0.03926042 0.03485128 0.03114187 0.02799222 0.02529554 0.01763055 0.01298162 0.00995329 0.00787205 0.00638084

1.5 1.51143911 1.52105263 1.52920354 1.53617021 1.54216867 1.55395683 1.5625 1.56888169 1.57377049 1.57759627 1.58064516 1.58311346 1.58513932 1.58682224 1.58823529 1.58943313 1.59045726 1.593361 1.59511699 1.59625877 1.59704251 1.59760359

2.59807621 2.94212205 3.32613496 3.75225662 4.22268193 4.73964614 6.25059468 8.10261339 10.3322275 12.9762868 16.0718414 19.6560699 23.7662369 28.4396661 33.7137226 39.6258016 46.2133204 53.513713 90.5930848 142.075496 210.359806 297.845382 406.931842

0.75 0.73843003 0.72850416 0.71994361 0.7125215 0.70605313 0.69313183 0.68359375 0.67637363 0.67078742 0.66638255 0.6628512 0.6599787 0.65761198 0.65563966 0.65397924 0.65256858 0.65136023 0.64792273 0.64583577 0.64447548 0.64354024 0.64286994

Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32

2.3 2.29880462 2.29522593 2.28928614 2.2810219 2.27048371 2.2577352 2.24285213 2.22592134 2.2070395 2.18631179 2.16385052 2.13977374 2.11420377 2.08726586 2.05908684 2.02979384

0 0.0021138 0.0084289 0.01886699 0.03329959 0.05155096 0.0734021 0.09859556 0.12684098 0.15782116 0.19119837 0.22662086 0.26372918 0.30216237 0.34156362 0.38158548 0.42189446

0 0.00091952 0.00367236 0.00824143 0.01459854 0.02270484 0.03251139 0.0439599 0.05698359 0.07150808 0.08745247 0.10473037 0.12325097 0.14292018 0.16364164 0.18531782 0.20785089

1.25517379 1.25484764 1.25387164 1.25225299 1.25000364 1.24714002 1.24368284 1.23965678 1.23509015 1.23001446 1.22446405 1.21847558 1.21208764 1.20534027 1.19827451 1.19093196 1.18335443

0 0.0018382 0.00733124 0.01641494 0.02898396 0.04489416 0.06396578 0.08598733 0.11072004 0.13790275 0.16725701 0.19849228 0.23131113 0.26541417 0.3005047 0.33629294 0.37249979

γ = 1.3

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407

Appendix G Table for Rayleigh Line Flow (continued) γ = 1.3 Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9 0.92 0.94 0.96 0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24 1.26

1.99951316 1.96836916 1.93648335 1.90397351 1.87095305 1.83753036 1.80380839 1.76988426 1.73584906 1.70178762 1.66777852 1.63389407 1.60020037 1.56675749 1.53361961 1.50083525 1.46844753 1.43649445 1.40500916 1.37402026 1.34355212 1.31362515 1.28425614 1.25545852 1.22724265 1.19961612 1.17258397 1.14614894 1.12031174 1.09507123 1.07042463 1.04636774 1.02289506 1 0.977675 0.95591169 0.93470098 0.91403319 0.89389817 0.87428537 0.8551839 0.83658267 0.81847039 0.80083565 0.78366702 0.766953 0.75068214

0.46217491 0.50213224 0.54149534 0.58001842 0.61748208 0.65369385 0.68848815 0.72172577 0.75329299 0.78310033 0.81108108 0.83718964 0.86139971 0.88370246 0.90410461 0.92262664 0.9393009 0.95416994 0.96728487 0.97870379 0.98849044 0.99671293 1.00344253 1.00875269 1.01271813 1.01541403 1.01691536 1.01729628 1.0166297 1.01498679 1.01243674 1.00904643 1.00488026 1 0.99446469 0.98833056 0.98165102 0.97447667 0.96685527 0.95883188 0.95044882 0.94174583 0.93276009 0.92352635 0.91407703 0.9044423 0.89465019

0.23114372 0.25510064 0.2796282 0.30463576 0.33003612 0.35574588 0.38168586 0.40778133 0.43396226 0.46016337 0.48632422 0.51238918 0.53830741 0.5640327 0.58952338 0.61474212 0.63965574 0.66423504 0.68845449 0.7122921 0.73572914 0.75874989 0.78134143 0.80349345 0.82519796 0.84644914 0.8672431 0.88757774 0.90745251 0.92686829 0.94582721 0.96433251 0.98238842 1 1.01717307 1.03391408 1.05023002 1.06612831 1.08161679 1.09670356 1.111397 1.12570564 1.13963816 1.15320334 1.16640999 1.17926693 1.19178297

1.1755835 1.16766021 1.15962477 1.15151625 1.14337242 1.13522948 1.12712199 1.11908271 1.11114253 1.10333045 1.09567349 1.08819679 1.08092355 1.07387513 1.0670711 1.06052928 1.05426585 1.04829544 1.04263122 1.03728499 1.03226728 1.02758745 1.0232538 1.01927364 1.01565341 1.01239874 1.00951458 1.00700521 1.0048744 1.00312541 1.00176111 1.000784 1.00019631 1 1.00019688 1.00078859 1.00177668 1.00316263 1.00494792 1.00713399 1.00972236 1.01271457 1.01611227 1.0199172 1.02413125 1.02875642 1.03379488

0.40886002 0.44512495 0.48106446 0.51646858 0.5511484 0.58493663 0.61768762 0.64927705 0.67960128 0.70857641 0.73613719 0.76223569 0.78683995 0.80993251 0.83150894 0.85157636 0.87015202 0.88726189 0.9029394 0.91722417 0.93016091 0.94179838 0.95218843 0.96138518 0.96944425 0.97642213 0.98237561 0.98736124 0.99143496 0.99465175 0.99706531 0.99872783 0.99968983 1 0.99970509 0.99884983 0.99747695 0.99562705 0.9933387 0.99064842 0.98759071 0.98419809 0.98050118 0.97652874 0.97230772 0.96786337 0.96321929

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408

Appendix G Table for Rayleigh Line Flow

(continued) γ = 1.3 Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

1.28 1.3 1.32 1.34 1.36 1.38 1.4 1.42 1.44 1.46 1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.64 1.66 1.68 1.7 1.72 1.74 1.76 1.78 1.8 1.82 1.84 1.86 1.88 1.9 1.92 1.94 1.96 1.98 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4

0.73484306 0.71942446 0.70441515 0.6898041 0.67558041 0.6617334 0.64825254 0.63512752 0.62234826 0.60990485 0.59778767 0.58598726 0.57449444 0.56330025 0.55239596 0.54177306 0.53142329 0.52133862 0.51151123 0.50193354 0.49259818 0.483498 0.47462608 0.46597567 0.45754026 0.44931353 0.44128933 0.43346174 0.42582499 0.41837351 0.4111019 0.40400492 0.39707751 0.39031476 0.38371193 0.37726441 0.37096774 0.31541415 0.27097078 0.23498161 0.20550393 0.18110236 0.1607043 0.14349888 0.12886598 0.11632612 0.10550459

0.88472671 0.87469593 0.86458011 0.85439979 0.84417389 0.83391979 0.82365345 0.81338949 0.80314126 0.79292095 0.78273964 0.77260741 0.76253335 0.7525257 0.74259185 0.73273845 0.72297143 0.71329605 0.70371699 0.69423835 0.68486373 0.67559622 0.66643851 0.65739286 0.64846116 0.63964497 0.63094553 0.62236378 0.61390043 0.60555591 0.59733046 0.58922411 0.58123671 0.57336795 0.56561738 0.55798439 0.55046826 0.48151266 0.42292895 0.37326257 0.33109783 0.29518259 0.26445694 0.23804268 0.21521947 0.1953987 0.17809949

1.20396687 1.21582734 1.22737296 1.23861223 1.24955353 1.26020508 1.27057497 1.28067114 1.29050134 1.30007319 1.3093941 1.31847134 1.32731197 1.33592288 1.3443108 1.35248226 1.36044362 1.36820106 1.37576059 1.38312805 1.39030909 1.39730923 1.40413379 1.41078794 1.41727672 1.42360498 1.42977744 1.43579866 1.44167308 1.44740499 1.45299854 1.45845776 1.46378653 1.46898864 1.47406775 1.47902738 1.48387097 1.5266045 1.56079171 1.58847568 1.61115082 1.62992126 1.64561207 1.65884702 1.67010309 1.67974914 1.68807339

1.03924899 1.04512128 1.05141447 1.05813151 1.06527554 1.07284995 1.08085835 1.08930459 1.09819278 1.10752727 1.11731267 1.12755385 1.13825596 1.14942441 1.16106488 1.17318334 1.18578604 1.19887952 1.21247059 1.22656638 1.24117429 1.25630204 1.27195763 1.28814937 1.30488589 1.32217612 1.3400293 1.35845499 1.37746307 1.39706373 1.41726751 1.43808527 1.45952817 1.48160776 1.50433589 1.52772477 1.55178695 1.83245772 2.19700355 2.66440918 3.25825213 4.00738323 4.94671565 6.11812114 7.57143378 9.36556261 11.5697154

0.95839751 0.95341856 0.94830154 0.9430642 0.93772304 0.93229332 0.92678918 0.92122371 0.91560897 0.90995608 0.90427529 0.89857601 0.89286688 0.88715582 0.88145007 0.87575625 0.87008039 0.86442798 0.85880398 0.8532129 0.84765881 0.84214538 0.83667588 0.83125326 0.82588014 0.82055882 0.81529135 0.81007953 0.80492489 0.79982877 0.79479232 0.78981649 0.78490205 0.78004965 0.77525977 0.77053277 0.76586889 0.7226877 0.68551266 0.65369637 0.62649467 0.60319921 0.58318505 0.56592061 0.55096185 0.53794111 0.526555

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409

Appendix G Table for Rayleigh Line Flow (continued) γ = 1.3 Mach Number

p/p*

T/T*

V/V*=ρ*/ρ

pt/pt*

Tt/Tt*

4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 12 14 16 18 20

0.084172 0.06865672 0.05703658 0.04811715 0.04112651 0.03554869 0.03102867 0.02731591 0.02422965 0.02163688 0.01943799 0.01755725 0.01222104 0.0089914 0.00689035 0.00544766 0.00441459

0.14346976 0.11784362 0.09840843 0.08334938 0.07146122 0.06192175 0.05415628 0.04775419 0.04241625 0.03792051 0.03409964 0.03082571 0.02150695 0.01584567 0.0121541 0.00961533 0.00779543

1.70448307 1.71641791 1.72535648 1.73221757 1.73759499 1.74188563 1.74536256 1.74821853 1.75059257 1.75258702 1.75427847 1.75572519 1.75982997 1.76231431 1.7639305 1.76504027 1.76583493

19.4370944 32.0625477 51.779242 81.7942018 126.417257 191.326139 283.869894 413.412737 591.720416 833.391101 1156.33277 1582.28905 4919.12069 13079.431 30834.3282 66110.8737 131309.281

0.5037036 0.48674538 0.47385798 0.46385743 0.45595363 0.44960572 0.44443465 0.44016904 0.43661079 0.43361278 0.43106399 0.42887944 0.42265842 0.4188769 0.4164102 0.41471339 0.41349686

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